We are providing you with Quant Section Mock for the upcoming SBI PO 2017 Prelim Exam. It contains 35 questions and time limit is 24 minutes.
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Review
Question 1 of 35
1. Question
1 points
A boat can row to a place 84 km away and come back in 26 hours. The time to row 21 km with the stream is same as the time to row 18 km against the stream. Find the speed of the stream.
If three dices are thrown simultaneously, what is the probability of having number 2 on third dice?
Correct
In all there sre 6^3 = 216 choices
On third dice 2 is fixed. On first 2 dices there can be any number
(1,1,2), (1,2,2)…..(1,6,2)
(2,1,2)…………..(2,6,2)
(3,1,2)…………..(3,6,2)
Like this, there are 36 favorable events
So probability = 36/216 = 1/6
Incorrect
In all there sre 6^3 = 216 choices
On third dice 2 is fixed. On first 2 dices there can be any number
(1,1,2), (1,2,2)…..(1,6,2)
(2,1,2)…………..(2,6,2)
(3,1,2)…………..(3,6,2)
Like this, there are 36 favorable events
So probability = 36/216 = 1/6
Unattempted
In all there sre 6^3 = 216 choices
On third dice 2 is fixed. On first 2 dices there can be any number
(1,1,2), (1,2,2)…..(1,6,2)
(2,1,2)…………..(2,6,2)
(3,1,2)…………..(3,6,2)
Like this, there are 36 favorable events
So probability = 36/216 = 1/6
Question 3 of 35
3. Question
1 points
6 years ago, three times the age of B was 2 more than the age if A that time. 6 years hence, twice age of B will be equal to A’s age that time. If the present age of C is the average of the present ages of A and B, find the age of C 5 years ago.
Correct
3 * (B-6) = 2 + (A-6)
2 * (B+6) = A + 6
Solve, A = 46, B = 20
So C’s present age = (46+20)/2 = 33
So 5 years ago = 28
Incorrect
3 * (B-6) = 2 + (A-6)
2 * (B+6) = A + 6
Solve, A = 46, B = 20
So C’s present age = (46+20)/2 = 33
So 5 years ago = 28
Unattempted
3 * (B-6) = 2 + (A-6)
2 * (B+6) = A + 6
Solve, A = 46, B = 20
So C’s present age = (46+20)/2 = 33
So 5 years ago = 28
Question 4 of 35
4. Question
1 points
Container A and B contains water and alcohol in the ratio 2 : 3 and 2 : 5 respectively. How much amount of mixture from container A should be mixed with 126 l of mixture from container B, so that the resultant mixture contains water and alcohol in the ratio 7 : 15?
Correct
Water in A = 2/5. Water in B = 2/7 . And in resultant water = 7/22
So by allegation method:
(x)…………….….(126)
2/5……………..…..2/7
………..7/22
5/(22*7)…………..9/(22*5)
Take ratio: 5/22*7 : 9/22*5
Gives 25 : 63
So x/126 = 25/63
Solve, x = 50 l
Incorrect
Water in A = 2/5. Water in B = 2/7 . And in resultant water = 7/22
So by allegation method:
(x)…………….….(126)
2/5……………..…..2/7
………..7/22
5/(22*7)…………..9/(22*5)
Take ratio: 5/22*7 : 9/22*5
Gives 25 : 63
So x/126 = 25/63
Solve, x = 50 l
Unattempted
Water in A = 2/5. Water in B = 2/7 . And in resultant water = 7/22
So by allegation method:
(x)…………….….(126)
2/5……………..…..2/7
………..7/22
5/(22*7)…………..9/(22*5)
Take ratio: 5/22*7 : 9/22*5
Gives 25 : 63
So x/126 = 25/63
Solve, x = 50 l
Question 5 of 35
5. Question
1 points
A and B start a business by investing Rs 2400 and Rs 2100 respectively. After 3 months A adds Rs 800 and B adds Rs 300 to their investments and third partner C joins them by investing Rs 3200. If share of B after a year is Rs 17,050 out of the total profit, find the profit of C.
Correct
Ratio shares of A : B : C is
2400*3 + 3200*9 : 2100*3 + 2400*9 : 3200*9
24*3 + 32*9 : 21*3 + 24*9 : 32*9
8 + 32 : 7 + 24 : 32
40 : 31 : 32
So
31/32 = 17050/x
Solve, x = Rs 17,600
Incorrect
Ratio shares of A : B : C is
2400*3 + 3200*9 : 2100*3 + 2400*9 : 3200*9
24*3 + 32*9 : 21*3 + 24*9 : 32*9
8 + 32 : 7 + 24 : 32
40 : 31 : 32
So
31/32 = 17050/x
Solve, x = Rs 17,600
Unattempted
Ratio shares of A : B : C is
2400*3 + 3200*9 : 2100*3 + 2400*9 : 3200*9
24*3 + 32*9 : 21*3 + 24*9 : 32*9
8 + 32 : 7 + 24 : 32
40 : 31 : 32
So
31/32 = 17050/x
Solve, x = Rs 17,600
Question 6 of 35
6. Question
1 points
Directions (6- 10): What approximate value should come in place of the question mark (?) in the following questions?
(Note: You are not expected to calculate the exact value.)
20.96% of 235.90 + 54.04% of 149.98 – 3.96% of 102.05 = ?
Correct
Incorrect
Unattempted
Question 7 of 35
7. Question
1 points
Directions (6- 10): What approximate value should come in place of the question mark (?) in the following questions?
(Note: You are not expected to calculate the exact value.)
Directions (6- 10): What approximate value should come in place of the question mark (?) in the following questions?
(Note: You are not expected to calculate the exact value.)
(3/4) + (2/3) + (1 1/3) + (2 2/3) = ?
Correct
Incorrect
Unattempted
Question 9 of 35
9. Question
1 points
Directions (6- 10): What approximate value should come in place of the question mark (?) in the following questions?
(Note: You are not expected to calculate the exact value.)
25.09 × 6.9 + 25.92% of 220.09 – √492 = ?
Correct
Incorrect
Unattempted
Question 10 of 35
10. Question
1 points
Directions (6- 10): What approximate value should come in place of the question mark (?) in the following questions?
(Note: You are not expected to calculate the exact value.)
45.009% of 640.08 – ? % of 959.99 = 48.06
Correct
Incorrect
Unattempted
Question 11 of 35
11. Question
1 points
Directions (11-15): What will come in place of question mark (?) in the following number series?
4, 5, 13, 23, 49, ?
Correct
×2 – 3, ×2 + 3, ×2 – 3, ×2 + 3, ….
Incorrect
×2 – 3, ×2 + 3, ×2 – 3, ×2 + 3, ….
Unattempted
×2 – 3, ×2 + 3, ×2 – 3, ×2 + 3, ….
Question 12 of 35
12. Question
1 points
Directions (11-15): What will come in place of question mark (?) in the following number series?
18, 9, 13.5, 33.75, ?
Correct
18*0.5, 9*1.5, 13.5*2.5, 33.75*3.5,
Incorrect
18*0.5, 9*1.5, 13.5*2.5, 33.75*3.5,
Unattempted
18*0.5, 9*1.5, 13.5*2.5, 33.75*3.5,
Question 13 of 35
13. Question
1 points
Directions (11-15): What will come in place of question mark (?) in the following number series?
20, 19, 36, 105, 416, ?
Correct
20*1 – 1, 19*2 – 2, 36*3 – 3, 105*4 – 4,
Incorrect
20*1 – 1, 19*2 – 2, 36*3 – 3, 105*4 – 4,
Unattempted
20*1 – 1, 19*2 – 2, 36*3 – 3, 105*4 – 4,
Question 14 of 35
14. Question
1 points
Directions (11-15): What will come in place of question mark (?) in the following number series?
13, 17, 24, 37, 59, ?
Correct
Double difference series
17 – 13 =4, 24 – 17 = 7, 37 – 24 = 13, 59-37 = 22
7 – 4 = 3, 13 – 7 = 6, 22 – 13 = 9
So next number 12+22+59
Incorrect
Double difference series
17 – 13 =4, 24 – 17 = 7, 37 – 24 = 13, 59-37 = 22
7 – 4 = 3, 13 – 7 = 6, 22 – 13 = 9
So next number 12+22+59
Unattempted
Double difference series
17 – 13 =4, 24 – 17 = 7, 37 – 24 = 13, 59-37 = 22
7 – 4 = 3, 13 – 7 = 6, 22 – 13 = 9
So next number 12+22+59
Question 15 of 35
15. Question
1 points
Directions (11-15): What will come in place of question mark (?) in the following number series?
Directions (16-20): Study the following table to answer the questions that follow:
How many marks did Kavya get in all the subjects?
Correct
Total marks of Kavya = 66/100 * 150 + 75 + 88/100 * 150 + 56/100 * 125 + 56/100 * 75 + 45
Incorrect
Total marks of Kavya = 66/100 * 150 + 75 + 88/100 * 150 + 56/100 * 125 + 56/100 * 75 + 45
Unattempted
Total marks of Kavya = 66/100 * 150 + 75 + 88/100 * 150 + 56/100 * 125 + 56/100 * 75 + 45
Question 17 of 35
17. Question
1 points
Directions (16-20): Study the following table to answer the questions that follow:
Marks obtained by Medha in Hindi is what percent of marks obtained by Sheena in the same subject?
Correct
Required % = 88/76 * 100 = 115.8%
Incorrect
Required % = 88/76 * 100 = 115.8%
Unattempted
Required % = 88/76 * 100 = 115.8%
Question 18 of 35
18. Question
1 points
Directions (16-20): Study the following table to answer the questions that follow:
what are the average marks obtained by all students together in Computer?
Correct
Average marks in Computer = (88+84+78+96+68+50)/6 = 464/6%
Required average marks = 464/600 * 150 = 116
Incorrect
Average marks in Computer = (88+84+78+96+68+50)/6 = 464/6%
Required average marks = 464/600 * 150 = 116
Unattempted
Average marks in Computer = (88+84+78+96+68+50)/6 = 464/6%
Required average marks = 464/600 * 150 = 116
Question 19 of 35
19. Question
1 points
Directions (16-20): Study the following table to answer the questions that follow:
Who has scored the highest total marks in all the subjects together?
Correct
Total by Medha = 568 – the highest
Incorrect
Total by Medha = 568 – the highest
Unattempted
Total by Medha = 568 – the highest
Question 20 of 35
20. Question
1 points
Directions (16-20): Study the following table to answer the questions that follow:
How many students have scored the highest marks in more than one subject?
Correct
Sheena in Physics and Chemistry
Medha in English and Computer
Incorrect
Sheena in Physics and Chemistry
Medha in English and Computer
Unattempted
Sheena in Physics and Chemistry
Medha in English and Computer
Question 21 of 35
21. Question
1 points
Directions (21-25): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 20x^{2} – 31x + 12 = 0,
II. 6y^{2} – 7y + 2 = 0
Directions (21-25): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 3x^{2} + 22 x + 24 = 0,
II. 3y^{2} – 10y + 3 = 0
Correct
3x^{2} + 22 x + 24 = 0
3x^{2} + 18x + 4x + 24 = 0
So x = -4/3, -6
3y^{2} – 10y + 3 = 0
3y^{2} – 9y – y + 3 = 0
So y = 1/3, 3
Incorrect
3x^{2} + 22 x + 24 = 0
3x^{2} + 18x + 4x + 24 = 0
So x = -4/3, -6
3y^{2} – 10y + 3 = 0
3y^{2} – 9y – y + 3 = 0
So y = 1/3, 3
Unattempted
3x^{2} + 22 x + 24 = 0
3x^{2} + 18x + 4x + 24 = 0
So x = -4/3, -6
3y^{2} – 10y + 3 = 0
3y^{2} – 9y – y + 3 = 0
So y = 1/3, 3
Question 23 of 35
23. Question
1 points
Directions (21-25): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. x^{2} – x – 6 = 0,
II. 5y^{2} – 7y – 6 = 0
Correct
x^{2} – x – 6 = 0
x^{2} – 2x + 3x – 6 = 0
So x = -3, 2
5y^{2} – 7y – 6 = 0
5y^{2} – 10y + 3y – 6 = 0
So y = -3/5, 2
Incorrect
x^{2} – x – 6 = 0
x^{2} – 2x + 3x – 6 = 0
So x = -3, 2
5y^{2} – 7y – 6 = 0
5y^{2} – 10y + 3y – 6 = 0
So y = -3/5, 2
Unattempted
x^{2} – x – 6 = 0
x^{2} – 2x + 3x – 6 = 0
So x = -3, 2
5y^{2} – 7y – 6 = 0
5y^{2} – 10y + 3y – 6 = 0
So y = -3/5, 2
Question 24 of 35
24. Question
1 points
Directions (21-25): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 2x^{2} + 17x + 30 = 0,
II. 2y^{2} + 13y + 18 = 0
Directions (21-25): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
Directions (26-30): Study the following charts carefully to answer the questions that follow:
The pie chart shows the distribution of number of phones sold by a shop during different months.The table shows the respective ratio between the number of phones sold by shop of company A and company B in different months.
What is the respective ratio between the number of phones sold of company B during January and those sold during June of the same company?
Correct
Ratio is
17*7/15 : 9*16/16
17*7 : 9*15
119 : 135
Incorrect
Ratio is
17*7/15 : 9*16/16
17*7 : 9*15
119 : 135
Unattempted
Ratio is
17*7/15 : 9*16/16
17*7 : 9*15
119 : 135
Question 27 of 35
27. Question
1 points
Directions (26-30): Study the following charts carefully to answer the questions that follow:
The pie chart shows the distribution of number of phones sold by a shop during different months.The table shows the respective ratio between the number of phones sold by shop of company A and company B in different months.
If 35% of the phones sold by company A during May were sold at a discount, home many phones of company A that month were sold without a discount?
Correct
Phones sold by company A during May = 7/15 * 12/100 * 45000
So sold without discount = 65/100 * 7/15 * 12/100 * 45000 = 1638
Incorrect
Phones sold by company A during May = 7/15 * 12/100 * 45000
So sold without discount = 65/100 * 7/15 * 12/100 * 45000 = 1638
Unattempted
Phones sold by company A during May = 7/15 * 12/100 * 45000
So sold without discount = 65/100 * 7/15 * 12/100 * 45000 = 1638
Question 28 of 35
28. Question
1 points
Directions (26-30): Study the following charts carefully to answer the questions that follow:
The pie chart shows the distribution of number of phones sold by a shop during different months.The table shows the respective ratio between the number of phones sold by shop of company A and company B in different months.
If the shopkeeper earned a profit of Rs 420 on each phone sold of company B during April, what was his total profit earned on the phones of that company during the same month?
Correct
Phones sold by B during April = 5/12 * 8/100 * 45000 = 1500
So profit earned = 420*1500 = Rs 6,30,000
Incorrect
Phones sold by B during April = 5/12 * 8/100 * 45000 = 1500
So profit earned = 420*1500 = Rs 6,30,000
Unattempted
Phones sold by B during April = 5/12 * 8/100 * 45000 = 1500
So profit earned = 420*1500 = Rs 6,30,000
Question 29 of 35
29. Question
1 points
Directions (26-30): Study the following charts carefully to answer the questions that follow:
The pie chart shows the distribution of number of phones sold by a shop during different months.The table shows the respective ratio between the number of phones sold by shop of company A and company B in different months.
The number of phones sold by company A during January is approximately what percent of the number of phones sold by company A during June?
Correct
Sold by A in January = 8/15 * 17/100 * 45000 = 4080
in June = 7/16 * 16/100 * 45000 = 3150
Required % = 4080/3150 * 100 = 130%
Incorrect
Sold by A in January = 8/15 * 17/100 * 45000 = 4080
in June = 7/16 * 16/100 * 45000 = 3150
Required % = 4080/3150 * 100 = 130%
Unattempted
Sold by A in January = 8/15 * 17/100 * 45000 = 4080
in June = 7/16 * 16/100 * 45000 = 3150
Required % = 4080/3150 * 100 = 130%
Question 30 of 35
30. Question
1 points
Directions (26-30): Study the following charts carefully to answer the questions that follow:
The pie chart shows the distribution of number of phones sold by a shop during different months.The table shows the respective ratio between the number of phones sold by shop of company A and company B in different months.
What is the total number of phones sold of company B during February and March together?
Correct
Sold by B in Feb = 5/9 * 22/100 * 45000 = 5500
Sold by B in March = 2/5 * 25/100 * 45000 = 45000
So total = 10,000
Incorrect
Sold by B in Feb = 5/9 * 22/100 * 45000 = 5500
Sold by B in March = 2/5 * 25/100 * 45000 = 45000
So total = 10,000
Unattempted
Sold by B in Feb = 5/9 * 22/100 * 45000 = 5500
Sold by B in March = 2/5 * 25/100 * 45000 = 45000
So total = 10,000
Question 31 of 35
31. Question
1 points
A work which is completed by 20 men in 8 days can be completed by 25 women 12 days. 16 men and 10 women start doing the work. After 3 days, they leave. If the remaining work is to be completed in 6 days by x number of men, find x.
Correct
20 men in 8 days so 16 men in 20 × 8/16 = 10 days and
25 women in 12 days so 10 women in 25 × 12/10 = 30 days
So in 3 days, they complete (1/10 + 1/30) × 3 = 2/5
So remaining work = 1 – 2/5 = 3/5
20 m 1 work in 8 days and x men 3/5 work in 6 days
So 20 × 8 × 3/5 = x × 6 × 1
So, x = 16 men
Incorrect
20 men in 8 days so 16 men in 20 × 8/16 = 10 days and
25 women in 12 days so 10 women in 25 × 12/10 = 30 days
So in 3 days, they complete (1/10 + 1/30) × 3 = 2/5
So remaining work = 1 – 2/5 = 3/5
20 m 1 work in 8 days and x men 3/5 work in 6 days
So 20 × 8 × 3/5 = x × 6 × 1
So, x = 16 men
Unattempted
20 men in 8 days so 16 men in 20 × 8/16 = 10 days and
25 women in 12 days so 10 women in 25 × 12/10 = 30 days
So in 3 days, they complete (1/10 + 1/30) × 3 = 2/5
So remaining work = 1 – 2/5 = 3/5
20 m 1 work in 8 days and x men 3/5 work in 6 days
So 20 × 8 × 3/5 = x × 6 × 1
So, x = 16 men
Question 32 of 35
32. Question
1 points
The circumference of a circle having radius equal to 35 cm is equal to the perimeter of a rectangle. If the area of rectangle is 2400 cm^{2}, find the length of rectangle.
Correct
2 × 22/7 × 35 = 2 (l + b)
so (l + b) = 110
also given, lb = 2400
So (l + 2400/l) = 110
So l^{2} – 110 l + 2400 = 0
So, l = 80 or 30. 30 not present in options.
Incorrect
2 × 22/7 × 35 = 2 (l + b)
so (l + b) = 110
also given, lb = 2400
So (l + 2400/l) = 110
So l^{2} – 110 l + 2400 = 0
So, l = 80 or 30. 30 not present in options.
Unattempted
2 × 22/7 × 35 = 2 (l + b)
so (l + b) = 110
also given, lb = 2400
So (l + 2400/l) = 110
So l^{2} – 110 l + 2400 = 0
So, l = 80 or 30. 30 not present in options.
Question 33 of 35
33. Question
1 points
A trader gets a profit of 30% on buying a kg of wheat. By mistake, he sold 1200 grams of wheat at the price of 1 kg. Find his net profit/loss%.
Correct
He sold 1200 grams of wheat at the price of 1 kg, means he incurs a loss here.
That loss% = (1200-1000)/1200 * 100 = 50/3%
Now there was a profit of 30% while buying and now a loss of 50/3% while selling. So successive method can be used here:
Profit/loss% = 30 + (-50/3) + (30)(-50/3)/100 = + 25/3%
Incorrect
He sold 1200 grams of wheat at the price of 1 kg, means he incurs a loss here.
That loss% = (1200-1000)/1200 * 100 = 50/3%
Now there was a profit of 30% while buying and now a loss of 50/3% while selling. So successive method can be used here:
Profit/loss% = 30 + (-50/3) + (30)(-50/3)/100 = + 25/3%
Unattempted
He sold 1200 grams of wheat at the price of 1 kg, means he incurs a loss here.
That loss% = (1200-1000)/1200 * 100 = 50/3%
Now there was a profit of 30% while buying and now a loss of 50/3% while selling. So successive method can be used here:
Profit/loss% = 30 + (-50/3) + (30)(-50/3)/100 = + 25/3%
Question 34 of 35
34. Question
1 points
In a class, average age of 30 students is 18 years. If the age of 2 more students is taken into consideration, then the average of all students gets increase by 1. Find the average of the ages of those 2 students.
Correct
30 students – 18
32 students – 19
So total age of those 2 students = 30×1 + 19×2 = 68
So average = 68/2 = 34
Incorrect
30 students – 18
32 students – 19
So total age of those 2 students = 30×1 + 19×2 = 68
So average = 68/2 = 34
Unattempted
30 students – 18
32 students – 19
So total age of those 2 students = 30×1 + 19×2 = 68
So average = 68/2 = 34
Question 35 of 35
35. Question
1 points
A person invests Rs 5800 in two schemes in such a way that he after 5 years and 4 years, he gets the same interest on respective parts. If the schemes offer 12% and 14% rate of interest respectively, what is the amount invested at 12% rate of interest?
Correct
Shortcut when interests are equal after t1 and t2 years:
1/(t1*r1) : 1/(t2*r2)
1/(5*12) : 1/(4*14)
14 : 15
So amount invested at 12% rate of interest = (14/29) * 5800 = 2800
Incorrect
Shortcut when interests are equal after t1 and t2 years:
1/(t1*r1) : 1/(t2*r2)
1/(5*12) : 1/(4*14)
14 : 15
So amount invested at 12% rate of interest = (14/29) * 5800 = 2800
Unattempted
Shortcut when interests are equal after t1 and t2 years:
1/(t1*r1) : 1/(t2*r2)
1/(5*12) : 1/(4*14)
14 : 15
So amount invested at 12% rate of interest = (14/29) * 5800 = 2800
mam some quadratic equations sums comprises of huge numbers andd factoring them means break them into prime and consume them takes lots of time is there any otherway for solving it
Yes there is a formula for calculating roots.
Suppose equation is ax^2 + bx + c
So its roots are [-b +- root (b^2 – 4ac)]/2a
+- means ek root k liye plus hoga, and second k liye minus.
But exam me se 5 questions me se 4 questions ki equations me jada bade factors nhi aate. To 5 me se 4 b attempt karoge to ok hai. Kabi kabi to all 5 are easy.
Is formula se dekho jaldi solve hota hai to use it, otherwise leave
Woah! I’m really enjoying the template/theme of this website.
It’s simple, yet effective. A lot of times it’s hard to get that “perfect balance” between usability and visual appearance.
I must say you have done a very good job with this.
Additionally, the blog loads very quick for me on Firefox.
31 recheck pls
crct he
thanks
18 of 35 questions answered correctly
Your time: 00:23:41
You have reached 18 of 35 points, (51.43)
Madam March month current affairs in PDF format
@shubhra madam
Yes check in downloads section in menu
NO4………….How much amount of mixture from container A should be mixed with 126 l of mixture from container B…ye line samjh mai nahi araha..
Container A me se kitna amount , container B k 126 l k sath mix krna chaiye?
Hn
Thqnk u
sir give me all concepts of RANKING test reasoning section ka?? plzzzzzz
19 of 35 questions answered correctly
Your time: 00:24:00
Thanks alot ?
explain q no 7 please
18^2 – 14^2 = 128
(2350+80)/x = 2430/x
so 128 + 2430/x = 230
x = 2430/102 = 24
Hlo frnds !! Is der any free full length mock test for SBI po,if der.pls send me dat link.my mail id swathimailapalli221995@gmail.com.thanks in advance..
We will now add some full length mock test as SBI PO pre is near now
Check here tomorrow
mam some quadratic equations sums comprises of huge numbers andd factoring them means break them into prime and consume them takes lots of time is there any otherway for solving it
Yes there is a formula for calculating roots.
Suppose equation is ax^2 + bx + c
So its roots are [-b +- root (b^2 – 4ac)]/2a
+- means ek root k liye plus hoga, and second k liye minus.
But exam me se 5 questions me se 4 questions ki equations me jada bade factors nhi aate. To 5 me se 4 b attempt karoge to ok hai. Kabi kabi to all 5 are easy.
Is formula se dekho jaldi solve hota hai to use it, otherwise leave
and mam have problem in like
1+x/4 like that some pllz provide quises on that time and is they come in exam like instalment i never hear it come
Yes installment k kabi nhi aae phle.
N 1+x/4 means ?
I dint get
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