Hello and welcome to exampundit. So here is a set of SBI PO Prelims Quantitative Aptitude Quiz 2017.
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.
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
(B) 100ml
(C) 200ml
(D) 150ml
(E) 220ml
2. 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?
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
6litre
(B) 6litre, 6litre
(C) 6litre, 5litre
(D) 5litre, 5litre
(E) none of these
3. 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?
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
(B) 45min
(C) 60min
(D) 20min
(E) 40min
4. 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?
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
(B) Tia
(C) Lovely
(D) Binny
(E) Gita
5. 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%?
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%
(B) 25%
(C) 13.67%
(D) 27.5%
(E) 33.33%
6. 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?
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
(B) 4/5
(C) 3/7
(D) 1/18
(E) 2/17
7. 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?
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
(B) 22years
(C) 21years
(D) 19years
(E) 13years
8. Area
of a Rhombus of perimeter 56 cms is 100 sq cms. Find the sum of the lengths of
its diagonals?
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
(B) 31.20cm
(C) 34.40cm
(D) 27.60cm
(E) 24.40cm
Answers with Solutions – Tap to show
1.A
Let initial
quantity of milk be 10x
After adding water,
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
quantity of milk be 10x
After adding water,
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
2. B
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
(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
3. C
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.
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.
4. D
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.
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.
5. E
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%
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%
6. A
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
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
7. B
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
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
8. C
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 .
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 .
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