# Mixed Quantitative Aptitude Questions Set 34

Quantitative Aptitude Questions for IBPS RRB/PO/Clerk, SBI PO, NIACL, NICL, RBI Grade B/Assistant, BOI, Bank of Baroda and other competitive exams

1. There was a book exhibition in an auditorium. On the first day 14 persons visited the exhibition , on the second day 12 persons and on the third day only 10 persons visited the exhibition . The ratio of admission fees collected from each of them on these days was 2 : 3 :5 resp. If the total amount collected on these three days was Rs. 4560 , what amount was collected on the first day ?
A) Rs.1400
B) Rs.1120
C) Rs.4500
D) Rs.2457
E) Rs.2450
Â Â Option B
Solution:
Ratio of amount
collected = (14*2) : (12*3) : (10*5) =Â  28 : 36 : 50 = 14 : 18 : 25
sum of ratios = 14 + 18 + 25 = 57
Therefore ,
Amount collected on day one = (14/57) * 4560 = Rs.1120
2. An aeroplane first flew with a speed of 440km/h and covered a certain distance .It still had to cover 770km less than what it had already covered but it flew with a speed of 660km/h . The average speed for the entire flight was 550km/h.Find the total dist. covered.
A) 1578km
B)Â 3245km
C) 4500km
D) 2750km
E) 2457km
Â Â Option D
Solution:

Let the plane covers x km with 440kmph and (x-770)km at a speed of 660kmph.
So, total dist. = (2x-770)km at a speed of 500kmph
Avg. speed = (Total dist./Total time)
=> 550 = (2x – 770)/[(x/440) +( (x-770)/660)]
=> x = 1760
Therefore ,
Total dist. covered = 2x â€“ 770 = 2750km
3. The length and breadth of the floor of a room are 20 feet and 10 feet resp. Square tiles of 2 feet length of three different colors are to be laid on the floor. Black tiles are laid in the first row on all sides. If white tiles are laid in the (1/3)rd of the remaining and blueÂ  tiles in the rest, how many blue tiles will be there ?
A) 10
B)Â 16
C) 20
D) 22
E) 18
Â Â OptionÂ  B
Solution:
Area of the floor = 20 * 10 = 200 sq. ft.
Area covered by black tiles = 2(20 + 20 + 6 + 6) = 104 sq. ft.
Remaining space = 200 â€“ 104 = 96 sq. ft.
Area covered by white tiles = (1/3)*96 = 32 sq. ft.
Area covered by blue tiles = 96 â€“ 32 = 64 sq. ft.
Area of 1 tile = (2)^2 = 4 sq. ft.
No. of blue tiles Â = 64/4 = 16Â
4. Romil had Rs. 4200. He invested some of it in scheme A for 4 years and rest of the money he invested in scheme B for two years. Scheme A offers simple interest at a rate of 22% pa and scheme B offers compounded interest at a rate of 10% pa. If the interest received from scheme A is Rs.1516 more than the interest received from scheme B, what was the sum invested by her in scheme A ?
A) Rs.1570
B)Â Rs.1245
C) Rs.2560
D) Rs.2200
E) Rs.1750
Â Â Option D
Solution:
Amount invested in scheme A = Rs. x
Amount invested in scheme B = Rs.(4200 – x)
Now,
(x*22*4)/100 â€“ (4200 – x)[(1 + (10/100)^2)] = 1516
=> (88x/100) â€“ (4200-x)[(121/100)-1] = 1516
=>(88x/100) â€“ (4200-x)(21/100) = 1516
=>(109x/100) = 1516 + 882
=> x = Rs.2200
5. A bottle contains (3/4) of milk and the rest water . How much of the mixture must be taken away and replaced by an equal quantity of water so that the mixture has half milk and half water ?
A) 15%
B)Â 10%
C) 33(1/3)%
D) 25%
E) 22%
Â Â Option C
Solution:
Let the mixture be 100 units
Milk = (3/4)*100 = 75
ratio of milk and water = 3 : 1
water = (1/4)*100 = 25
Let x litre of mixture is taken out then the remaining quantity of milk left = [3 – (3x/4)]
and water left =Â  [1 â€“ (x/4)] + x
Given , 3 â€“ (3x/4)Â  = 1 â€“ (x/4) + x
=> x = 4/3
Req. percentage = (1/3)*100 = 33(1/3)%
6. There are two companies P and Q quote for a tender . On the tender opening day . P realizes that the two quotes are in the ratio 7 : 4 and hence decreases its price during negotiations to make it Rs.1,00,000 lower than Qâ€™s quoted price. Q then realizes that the final quotes of the two were in the ratio 3 : 4 . By how much did P decrease its price in order to win the bid?
A) 7lakhs
B)Â 4lakhs
C) 8lakhs
D) 3lakhs
E) 6lakhs
Â Â Option B
Solution:
Initially, Pâ€™s quote be 7x and Qâ€™s quote be 4x.
new ratio changes = 3 : 4
change is decrease = 4lakhs
7. The number of copies circulated for a first-rank newspaper is 30,15,450 in the world. If the number of copies circulated for the second rank newspaper Â is less by 11,24,000 of the first rank newspaper .Find how many copies of second rank newspaper are circulated in the world?
A) 15,55,530
B)Â 18,91,450
C) 16,52,000
D) 14,45,500
E) 11,15,000
Â Â Option B
Solution:
Number of copies of second rank = 30,15,450 â€“ 11,24,000 = 18,91,450
8. Two pipes can separately fill a tankÂ  in 10hrs. and 30 hrs. resp. Both the pipes are opened to fill the tank but when the tank is (1/2) full a leak developes in the tank through which (1/2) of the water supplied by both the pipes per hour leak out .What is the total time taken to fill the tank?
A) 9hrs
B)Â 10.5hrs
C) 15hrs
D) 11.25hrs
E) 12hrs
Â Â Option D
Solution:
(10*30)/(10+30) = 300/40 = 7.5 hrs.
Therefore ,
(1/2) of the tank if filled in (7.5/2) = 3.75 hrs.
Now, (1/2) of the supplied water leaks out
Filler pipes only = 1 â€“ (1/2) =1/2
Work Â 3.75 hrs will beÂ  completed in = 3.75 / (1/2) = 7.5 hrs.
Total time =Â  Â 7.5 + 3.75 =Â  11.25 hrs.
9. A family consists of a grandfather , 5 sons and daughters and 8 granchildren . They are to be seated in a row for dinner . The grandchildren wish to occupyÂ  4 seats at each end and the grandfather refuses to have a grandchild on either side of him.Find the number of ways in which the family can be made to sit .
A) 1254755
B)Â 1545000
C) 1245700
D) 15784582
E) 19353600
Â Â Option E
Solution:
Total members in a family = 1 + 5 + 8 = 14
The grandchildren occupy the 4 seats on either side of the table = 8P4 *Â  4! = 8!
The grandfather occupy a seat in 4 ways.
The remaining seats can be occupied in 5! = 120 ways
Required ways = 8! * 480 = 19353600
Â  Â Â
10. There are 8 blue balls , 4 green balls and 5 white balls in a bag. 5 balls are chosen at random . What is the probability of their being 2 blue balls , 1 green ball and 2 white balls?
A)Â  199/1457
B)Â 200/1547
C) 280/1547
D) 280/1457
E) 110/1457
Â Â Option C
Solution:
Total possible outcomes = 17C5 = 6188
Total favourable outcomesÂ  = 8C2 * 4C1 * 5C2 = 1120
Required probability = 1120/6188 = 280/1547

## 11 Thoughts to “Mixed Quantitative Aptitude Questions Set 34”

1. Aspirant !!

8/10
Ty az ðŸ™‚

thank u so much banya mam…mam check the spd question in question it should be 500 intsead of 550..:)

Apka 2nd question allegation se ban raha hai?
440…………….660
……….500.
160……………60
8……………….3 (time ratio)
440*8+660*3 == Total distance hone chaiye na? Ans to nahi aa rahe. Please explain kardo Sir

3. Diya

in Q3
Area covered by black tiles = 2(20 + 20 + 6 + 6) = 104 sq. ft.
how to find this?

1. Banya

Ok..
After ignoring the black tiles,the area of the floor =(20 – 2 – 2)*(10 – 2 -2 )—————-[ Use this ]
= 104sq. ft.

4. Purvi !!

thanku mam ðŸ™‚

5. Jeetesh Chandra

ty..

6. Marry..

thank u so much……

7. subhain