# SBI Clerk 2018: Quantitative Aptitude Quiz for Prelims – 15

0
(0)

Hello and welcome to exampundit. Here is a set of Quantitative Aptitude Quiz on Mixed Problems for Prelims exam of SBI Clerk 2018.

1. A milkman adds 500 ml of water to each litre of milk he has in a container. He sells 30 litre of mixture from container and adds 10 litre milk in the remaining. The ratio of milk and water in the final mixture is 11:5. Find the initial quantity of milk in the container.

(A) 120ml
(B) 100ml
(C) 200ml
(D) 150ml
(E) 220ml

Option: A

Explanation: Let initial quantity of milk be 10x
total quantity =15x
quantity of milk =10x
After selling 30 litre of mixture and adding 10 litre milk,
total quantity =(15x−30+10)=(15x−20)
quantity of milk =10x−(30×10x15x)+10=(10x−10)
10x−1015x−20=1111+5⇒x=12
Initial quantity of milk in the container
=10x=120 litre

1. A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?

(A) 5litre, 6litre
(B) 6litre, 6litre
(C) 6litre, 5litre
(D) 5litre, 5litre
(E) none of these

Option: B

Explanation: Let x and (12-x) litres of milk be mixed from the first and second container respectively
Amount of milk in x litres of the first container = .75x
Amount of water in x litres of the first container = .25x
Amount of milk in (12-x) litres of the second container = .5(12-x)
Amount of water in (12-x) litres of the second container = .5(12-x)
Ratio of water to milk = [.25x + .5(12-x)] : [.75x + .5(12-x)]=3:5 ⇒(.25x+6−.5x)/(.75x+6−.5x)=3/5
⇒(6−.25x)/(.25x+6)=3/5
⇒30−1.25x=.75x+18
⇒2x=12⇒x=6
Since x = 6, 12-x = 12-6 = 6  Hence 6 and 6 litres of milk should mixed from the first and second container respectively

1. Two friends A and B leave City P and City Q simultaneously and travel towards Q and P at constant speeds. They meet at a point in between the two cities and then proceed to their respective destinations in 54 minutes and 24 minutes respectively. How long did B take to cover the entire journey between City Q and City P?

(A) 90min
(B) 45min
(C) 60min
(D) 20min
(E) 40min

Option: C

Explanation: Let us assume Car A travels at a speed of a and Car B travels at a speed of b. Further, let us assume that they meet after t minutes.
Distance traveled by car A before meeting car B = a * t. Likewise distance traveled by car B before meeting car A = b * t.
Distance traveled by car A after meeting car B = a * 54. Distance traveled by car B after meeting car A = 24 * b.
Distance traveled by car A after crossing car B = distance traveled by car B before crossing car A (and vice versa).
=> at = 54b ———- (1)
and bt = 24a ——– (2)
Multiplying equations 1 and 2
we have ab * t2 = 54 * 24 * ab
=> t2 = 54 * 24
=> t = 36
So, both cars would have traveled 36 minutes prior to crossing each other. Or, B would have taken 36 + 24 = 60 minutes to travel the whole distance.

1. Tia, Mina, Gita, Lovely and Binny are 5 sisters, aged in that order, with Tia being the eldest. Each of them had to carry a bucket of water from a well to their house. Their buckets’ capacities were proportional to their ages. While returning, equal amount of water got splashed out of their buckets. Who lost maximum amount of water as a percentage of the bucket capacity?

(A) Mina
(B) Tia
(C) Lovely
(D) Binny
(E) Gita

Option: D

Explanation: Tia is the older and Binny is the youngest.
So, Binny’s bucket would have been the smallest.
Each sister lost equal amount of water.
As a proportion of the capacity of their buckets Binny would have lost the most.

1. What is the maximum percentage discount that Sundarnath can offer on his marked price so that he ends up selling at no profit or loss, if he had initially marked his goods up by 50%?

(A) 67.67%
(B) 25%
(C) 13.67%
(D) 27.5%
(E) 33.33%

Option: E

Explanation: Let the cost price of the goods to be 100x
he had initially marked his goods up by 50%.
Therefore, a 50% markup would have resulted in his marked price being 100x + 50% of 100x = 100x + 50x = 150x.
He finally sells the product at no profit or loss.
i.e., he sells the product at cost price, which in this case is 100x.
Therefore, he offers a discount of 50x on his marked price of 150x.
Hence, the % discount offered by him= Discount/MarkedPrice×100=50/150×100DiscountMarked Price×100=50/150×100= 33.33%

1. In a game there are 70 people in which 40 are boys and 30 are girls, out of which 10 people are selected at random. One from the total group, thus selected is selected as a leader at random. What is the probability that the person, chosen as the leader is a boy?

(A) 4/7
(B) 4/5
(C) 3/7
(D) 1/18
(E) 2/17

Option: A

Explanation: The total groups contains boys and girls in the ratio 4:3
If some person are selected at random from the group, the expected value of the ratio of boys and girls will be 4:3
If the leader is chosen at random from the selection, the probability of him being a boy = 4/7

1. If a sum of money grows to 144/121 times when invested for two years in a scheme where interest is compounded annually, how long will the same sum of money take to treble if invested at the same rate of interest in a scheme where interest is computed using simple interest method?

(A) 18years
(B) 22years
(C) 21years
(D) 19years
(E) 13years

Option: B

Explanation: The sum of money grows to 144/121 times in 2 years.
If P is the principal invested, then it has grown to 144/121 P in two years when invested in compound interest.
In compound interest, if a sum is invested for two years, the amount is found using the following formula
A=(1+R/100)² P in this case.
=>(1+R/100)²=144/121 =>(1+R/100)²=(12/11)²=>R=100/11
If r =100/11% , then in simple interest the time it will take for a sum of money to treble is found out as follows:
Let P be the principal invested. Therefore, if the principal trebles = 3P, the remaining 2P has come on account of simple interest.
Simple Interest =PNR/100 , where P is the simple interest, R is the rate of interest and ‘N’ is the number of years the principal was invested.
Therefore, 2P =PN×100/11×100  => 2 =N/11 or N = 22 years

1.  Area of a Rhombus of perimeter 56 cms is 100 sq cms. Find the sum of the lengths of its diagonals?

(A) 29.80cm
(B) 31.20cm
(C) 34.40cm
(D) 27.60cm
(E) 24.40cm

Option: C

Explanation: Perimeter = 56. Let the side of the rhombus be “a”, then 4a = 56 => a =14.
Area of Rhombus = Half the product of its diagonals. Let the diagonals be d1 and d2 respectively.
1/2×d1× d2 = 100 => d1×d2 = 200.
By Pythagoras theorem, (d1)² + (d2)²= 4a² => (d1)² + (d2)² = 4×196 = 784.
(d1)² + (d2)² + 2d1× d2 = (d1+ d2)² = 784 +2×200 = 1184 => (d1+ d2) = √1184 = 34.40
Therefore, sum of the diagonals is equal to 34.40 cm .

Regards

Team Exampundit

Average rating 0 / 5. Vote count: 0

No votes so far! Be the first to rate this post.

We are sorry that this post was not useful for you!

Let us improve this post!

Tell us how we can improve this post?