Directions(1-5): Find the values of x and y and then compare the values and choose a correct option.
- I.4x^2 – 23x – 6 = 0
II.4y^2 + 19y + 12 = 0y > xNo relationx > yy >= xx >= yOption C
I.4x^2 – 23x – 6 = 0
=>4x^2 – 24x + x – 6 = 0
=>(x-6)(4x + 1) = 0
=>x = 6,-1/4
II.4y^2 + 19y + 12 = 0
=>4y^2 + 16y + 3y + 12 = 0
=>(y+4)(4y+3) =0
=>y = -4,-3/4
x > y - I.x^2 -14x + 48= 0
II.y^2 – 2y – 63 = 0
x > yx >= yy > xNo relationy >= xOption D
I.x^2 -14x + 48= 0
=>x^2 – 8x – 6x + 48 = 0
=>(x-6)(x-8) = 0
=>x = 6,8
II.y^2 – 2y – 63 = 0
=>y^2 -9y + 7y – 63 = 0
=>(y+7)(y-9) = 0
=>y = -7,9
No relation - I.30x^2 – 49x + 20 = 0
II.63y^2 – 64y + 16 = 0x >= yx > yy > xy >= xNo relationOption B
I.30x^2 – 49x + 20 = 0
=>30x^2 – 24x – 25x +20 = 0
=>(6x-5)(5x-4) = 0
=>x = 5/6,4/5
II.63y^2 – 64y + 16 = 0
=>63y^2 -28y – 36y + 16 = 0
=>(7y-4)(9y -4) = 0
=>y = 4/9,4/7
x > y - I.35x^2 – 43x + 12 = 0
II.42y^2 – 65y + 25 = 0
x>= yy >= xNo relationy > xx > yOption C
I.35x^2 – 43x + 12 = 0
=>35x^2 – 15x – 28x + 12 = 0
=>(5x-4)(7x-3) = 0
=>x = 4/5,3/7
II.42y^2 – 65y + 25 = 0
=>42y^2 – 30y – 35y + 25 = 0
=>(6y -5)(7y -5) = 0
=>y = 5/6,5/7
No relation - I.x^2 + x – 42 = 0
II.y^2 + 15y + 56 = 0
y >= xx >= yy > xx > yNo relationOption B
I.x^2 + x – 42 = 0
=>x^2 + 7x – 6x – 42 = 0
=>(x-6)(x+7) = 0
=>x = 6,-7
II.y^2 + 15y + 56 = 0
=>y^2 + 8y + 7y + 56 = 0
=>(y+8)(y+7) = 0
=>y = -8,-7
x >= y - A vessel contains a mixture of milk and water in the ratio of 14 : 3. Now, 25.5 liters of the mixture is taken out from the vessel and 2.5 litres of pure water and 5 litres of pure milk is added to the mixture. If the resultant mixture contains 20% water, what was the initial quantity of mixture in the vessel before the replacement? (in litres)
7560906880Option D
Let Milk = 14𝑥
Water =3𝑥
Now,
[14𝑥+5]/[3𝑥+2.5]=4/1
14𝑥+5=12𝑥+10
=>2𝑥=5
=> 𝑥=2.5
Initial quantity =17×2.5+25.5=68 - In a circular race of 900m length, A and B start with speeds 27 km/hr and 36 km/hr starting at the same time from the same point. When will they meet for the first time at the starting point when running in opposite direction?
8 min.4 min.6 min.9 min.5 min.Option C
Time taken by A to reach start point =900/(27×5)×18 = 120 sec
Time taken by B to reach start point =900/(36×5)×18 =90 sec.
LCM of 90 and 120 = 30×3×4 = 360 sec.
Required time =360/60=6 min. - After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago. The difference between the ages of the replaced and the new members is :
1013151112Option C
Let present average =𝑥 year
Total age =5𝑥 years According to questions 5𝑥−𝑦+𝑧=5𝑥−15
𝑦-> Old member, 𝑧-> New member
−𝑦+𝑧=−15
=>𝑦−𝑧=15 - A watch dealer usually sells watches for Rs.2350 per watch. Once he gave two successive discounts of 15% and 25% while selling a watch to a customer. But he charged an additional 8% on the net sale price from the customer. By what percent is the new selling price less than the original selling price?
15.15%31.15%14.27%22.40%20.25%Option B
Let usually S.P. = 100
After discounts price =85/100×75/100×108/100×100 = 68.85%
Requried % = 100−68.85 =31.15% - How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27kg of sugar costing Rs.7 per kg so that there may be gain of 10% by selling the mixture at Rs.9.24 per kg?
48 kg55 kg60 kg50 kg63 kgOption E
Mean price =10/110×9.24
=10×0.84
= 8.49 ====7
===8.4
1.4===0.6
Ratio = 1.4/0.6 = 7/3
Required Quantity = 27*7/3 = 63 kg - The average price of 10 books is Rs.12 while the average price of 8 of these books is Rs.11.75. Of the remaining two books, if the price of one book is 60% more than the price of the other, what is the price of each of these one book?
1410151316Option E
Sum of price of the remaining two
Books = 12 × 10 – 11.75 × 8
= 26
Let cost of First book
=> 260𝑥/100=26
x = 10
Price of second book = 10 + 6 = 16 - A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office?
7 km5 km6 km4 km8 kmOption C
Average speed =2×3×2/(2+3)=12/5 km/hr.
Total time taken =5 hoursDistance travelled =12/5×5=12 km
Therefore, distance between his house and office
=12/2=6 km - A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
30(3/10)%33(2/13)%32(4/12)%40(1/13)%45(5/11)%Option E
Total runs scored = 110Total runs scored from boundaries and sixes
=3×4+8×6=60
Total runs scored by running between the wickets
=110−60=50
Required percent
=50/110×100=500/11= 45(5/11)% - A starts a business with a capital of Rs. 85,000. B joins in the business with Rs.42500 after some time. For how much period does B join, if the profits at the end of the year are divided in the ratio of 3 : 1?
97864Option C
Let B joined for xx months. Then85000×12:42500×x=3:1
=>850×12:425x=3:1
=>850×12×1=3×425x
=>850×4=425x
=>x=8 - Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increases by:
1/21/41/61/71/5Option D
Let the cost of the rental car =xWhen 8 persons share equally, share of one person =x/8
When 1 person withdraws and other 7 persons share equally, share of one person =x/7
Increase in the share =x/7−x/8=x/56
Required fraction =(x/56)/(x/8) = 1/7
- 567, 1140, 2290, ?, 9206, 18434
43204594442541054110Option B
×2 + 6,×2 + 10,× 2 + 14,× 2+ 18
? = 4594 - 17, 38, ?, 508, 2565, 15426
111120105123116Option D
×2+ 2^2,×3+3^2,× 4 + 4^2
? = 123 - 2, 60 ,10, 120 ,30, ?
250230240210220Option D
1^3+1, 4^3 – 4,2^3+2,5^3 –5,3^3+3,6^3 – 6
? = 210 - 6,15, 45, 157.5, ?, 2835
630600619635616Option A
×2.5,× 3,× 3.5,× 4
? = 630 - 210, 130, 60, ? ,6, 2
3040506080Option A
6^3−6,5^3+5,4^3−4,3^3+3,2^3−2,1^3+1
? = 30 - (60% of 40% of 3250 – 23*24) = ?
241228230212200Option B
(60% of 40% of 3250 – 23*24) = ?
=>60% of 1300 – 552 = ?
=>780 – 552 = ?
=>? = 228 - [(872/32+12)*4 – (774/43)] = ?
120144125139110Option D
[(872/32+12)*4 – (774/43)] = ?
=>[(27.25+12)*4 – 18] = ?
=>39.25*4 – 18 = ?
=>? = 139 - (26*7 + 29*4 – 32*5) = ?
156138152160144Option B
(26*7 + 29*4 – 32*5) = ?
=>182 + 116 – 160 = ?
=>? = 138 - 16 of 19 – 702/26 – 17*14 = ?
4045423930Option D
16 of 19 – 702/26 – 17*14 = ?
=>304 – 27 -238 = ?
=>? = 39 - 228/12 + 16*7 + 75 = ?
212200220206215Option D
228/12 + 16*7 + 75 = ?
=>19 + 112 + 75 = ?
=>? = 206 - 20% of 150 + 323 /17 + (?)^1/2 = 27% of 300
10101024112213251440Option B
20% of 150 + 323 /17 + (?)^1/2 = 27% of 300
=>30+19 + (?)^1/2 = 81
=>(?)^1/2 = 81 -49
=>(?)^1/2 = 32
=>? = 1024 - (81)^1/2 *(625)^1/2 + 1225 = (?)^2 – 150
5040603070Option B
(81)^1/2 *(625)^1/2 + 1225 = (?)^2 – 150
=>9*25 + 1225 = (?^2 ) – 150
=>?^2 – 150 = 225+1225
=>?^2 = 1450+150
=>? = 40 - [(256)^1/2 + 192 /8 ]% of 500 = ?^2 of 8
87654Option D
[(256)^1/2 + 192 /8 ]% of 500 = ?^2 of 8
=>?^2 * 8 = (16+24)% * 500
=>?^2 * 8 = 40% of 500
=>?^2 = 200/8
=>?^2 = 25
=>? = 5 - (576)^1/2 + (676)^1/2 *8 = ?% of 250
130100150160140Option D
(576)^1/2 + (676)^1/2 *8 = ?% of 250
=>?*2.5 = (24+26)*8
=>?*2.5 = 50*8
=>? = 400/2.5
=>? = 160 - 60% of 250 + 14^2 = ?^2 – (225)^1/2
1519171816Option B
60% of 250 + 14^2 = ?^2 – (225)^1/2
=>?^2 – 15 = 0.6*250 + 196
=>?^ 2= 150+196 +15
=>?^2 = 361
=>? = 19 - The total number of Amul stores opened in 2010 in first quarter and third quarter combined was what percentage of the total number of Amul stores opened in 2011 in first quarter and third quarter combined?
75%65%81%92%70%Option C
Total number of Amul stores opened in 2010 in first and third quarter = 140+160 = 300
Total number of Amul stores opened in 2011 in first and third quarter = 180+190 = 370
Required% = 300/370*100 = 81.08% == 81% - 30% of the total number of Amul stores opened in 2013 were about to shut down. What was the total number of Amul store opened in 2013 that were about to shut down?
255222220235240Option A
Total number of stores opened in 2013 = 260+150+230+210 = 850
Total number of stores about to shut down = 0.3*850 = 255 - What was the ratio of the total number of Amul store opened in third quarter in the given years combined and the total number of Amul store opened in fourth quarter in the given years combined ?
35:3633:3832:3539:3431:29Option D
Total number of Amul stores opened in third quarter in 4 years = 160+190+ 200+230 = 780
Total number of Amul stores opened in fourth quarter in 4 years = 120+160+190+210 = 680
Required ratio = 780:680 = 39:34 - What was the difference between the average of the number of Amul store opened in first quarter in 2010 and 2012 together and the average of the number of Amul store opened in fourth quarter in 2011 and 2013 together?
1614121018Option D
Average number of Amul store opened in first quarter in 2010 and 2012 together = (140+250)/2 = 195
Average number of Amul store opened in first quarter in 2011 and 2013 together = (160+210)/2 = 185
Required Difference = 195-185 = 10 - Find the total number of Amul stores opened 2009, if the number of Amul stores opened increased by % in 2010
as compared to the previous year.
395410400350448Option C
Total number of Amul stores opened in 2010 = 140+80+160+120 = 500
Total number of Amul stores opened in 2009 = 500/1.25 = 400
Directions(16-20): Find the missing term ‘?’ of the following series.
Directions(21-30):approximate value will come in place of question mark ‘?’ in the following questions.
Directions(31-35): Answer the following questions based on the given information given below:
The graph represents the number of Amul store opened in different quarters in different years.
