Best Quant Questions for CAT
Download important CAT Best Quant Questions with Solutions PDF based on previously asked questions in CAT exam. Practice Best Quant Questions with Solutions for CAT exam.
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Question 1: Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight-member joint family is nearest to[CAT 2007]
a) 23 years
b) 22 years
c) 21 years
d) 25 years
e) 24 years
Question 2: A student gets an aggregate of 60% marks in five subjects in the ratio 10 : 9 : 8 : 7 : 6. If the passing marks are 50% of the maximum marks and each subject has the same maximum marks, in how many subjects did he pass the examination?
a) 2
b) 3
c) 4
d) 5
Question 3: Product M is produced by mixing chemical X and chemical Y in the ratio of 5 : 4. Chemical X is prepared by mixing two raw materials, A and B, in the ratio of 1 : 3. Chemical Y is prepared by mixing raw materials, B and C, in the ratio of 2 : 1. Then the final mixture is prepared by mixing 864 units of product M with water. If the concentration of the raw material B in the final mixture is 50%, how much water had been added to product M?
a) 328 units
b) 368 units
c) 392 units
d) 616 units
e) None of the above
Question 4: An outgoing batch of students wants to gift PA system worth Rs.4200 to their school. If the teachers offer to pay 50% more than the students, and an external benefactor gives three times teachers. contribution, how much should the teachers donate?
a) 600
b) 840
c) 900
d) 1200
Instructions Answer the questions based on the following information. A salesman enters the quantity sold and the price into the computer. Both the numbers are two-digit numbers. But, by mistake, both the numbers were entered with their digits interchanged The total sales value remained the same, i.e. Rs. 1,148, but the inventory increased by 54.
Question 5: What is the actual quantity sold?
a) 28
b) 21
c) 82
d) 14
Instructions
Answer the following questions based on the following information.
A company purchases components A and B from Germany and USA respectively. A and B form 30% and 50% of the total production cost. Current gain is 20%. Due to change in the international scenario, cost of the German mark increased by 30% and that of USA dollar increased by 22%. Due to market conditions, the selling price cannot be increased by more than 10%.
Question 6: If the USA dollar becomes cheap by 12% over its original cost and the cost of German mark increased by 20%, what will be the gain? (The selling price is not altered.)
a) 10%
b) 20%
c) 15%
d) 7.5%
Question 7: A pharmaceutical company manufactures 6000 strips of prescribed diabetic drugs for Rs. 8,00,000 every month. In July 2014, the company supplied 600 strips of free medicines to the doctors at various hospitals. Of the remaining medicines, it was able to sell 4/5th of the strips at 25 percent discount and the balance at the printed price of Rs. 250. Assuming vendor’s discount at the rate of a uniform 30 percent of the total revenue, the approximate percentage profit / loss of the pharmaceutical company in July 2014 is:
a) 5.5 percent (profit)
b) 4 percent (loss)
c) 5.5 percent (loss)
d) None of the above
Question 8: The number of positive integers n in the range $12 \leq n \leq 40$ such that the product (n -1)*(n – 2)*…*3*2*1 is not divisible by n is
a) 5
b) 7
c) 13
d) 14
Question 9: If a/b = 1/3, b/c = 2, c/d = 1/2 , d/e = 3 and e/f = 1/4, then what is the value of abc/def ?
a) 3/8
b) 27/8
c) 3/4
d) 27/4
e) 1/4
Question 10: Let $x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$. Then x equals
a) 3
b) $(\sqrt{13} – 1)/2$
c) $(\sqrt{13} + 1)/2$
d) $\sqrt{13}$
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Answers & Solutions:
1) Answer (E)
Ten years ago, the total age of the family is 231 years.
Seven years ago, (Just before the death of the first person), the total age of the family would have been 231+8*3 = 231+24 = 255.
After the death of one member and the birth of a child, the total age is 195 years.
Four years ago, after the death of one member and birth of another child, the total age of the family is 195+24-60 = 159 years.
The current total age of the family is = 8×4 + 159 = 191 years
The average age is 191/8 = 23.875 years = 24 years (approx)
2) Answer (C)
Let’s say he scored marks as $10x,9x,8x,7x,6x$ or total of $40x$ which is 60% of total maximum marks(T).
$\frac{T \times 60}{100}=40x$
So T (total maximum marks)=$\frac{400x}{6}$
Or Individual max. marks = $\frac{T}{5}=\frac{80x}{6}$
Passing marks =50% of individual max. marks =$\frac{40x}{6}=6.66x$
Hence he scored more than passing marks in four subjects as $10x,9x,8x$ and $7x$
and failed in one subject as scoring $6x$ marks which is less than passing marks of $6.66x$
3) Answer (B)
Let the quantities of the chemicals X and Y, mixed to produce product M be $5c$ and $4c$ respectively.
X is prepared by mixing A and B in the ratio = 1 : 3
=> Quantity of B in X = $\frac{3}{4} \times 5c = \frac{15 c}{4}$
Y is prepared by mixing B and C in the ratio = 2 : 1
Quantity of B in Y = $\frac{2}{3} \times 4c = \frac{8 c}{3}$
Quantity of B in M = $\frac{15 c}{4} + \frac{8 c}{3} = \frac{77 c}{12}$
Now, 864 units of M was mixed with water to prepare the final mixture.
=> Total quantity of M = $9c = 864$ => $c = \frac{864}{9} = 96$
Concentration of raw material B in the final mixture is 50 %
=> Quantity of final mixture = $\frac{100}{50} \times \frac{77}{12} \times 96 = 1232$
$\therefore$ Quantity of water added to M = $1232 – 864 = 368$ units
4) Answer (C)
Let’s, say students contributed x rs.
So teachers contributed = $\frac{3x}{2}$
And external benefactor contributed = $\frac{9x}{2}$
Summation of all contribution = 4200 = $7x$
or $x=600$
Hence, teachers contributed = 900
5) Answer (A)
Total sales value = 1148
Now 1148 can be represented as $41 \times 28$ and $82 \times 14$ i.e. product of interchanged numbers.
As inventory is increased by 54, so quantity sold should be 28.
6) Answer (B)
Let the total production cost be 100.
Hence, selling price is 120.
Price of German component A is 30 and the price of the US component B is 50
After change in exchange rate, price of German component is 30*1.2 = 36
and price of US component is 50*0.88=44
Total increase equals (36+44)-(30+50) = 0
Hence, the total production cost did not change.
As the selling price also did not change, the gain percentage equals 20%
7) Answer (C)
It is given that a total of 6000 strips are manufactured out of which the company supplied 600 strips of free medicines to the doctors.
Hence, the number of strips which were sold = 6000 – 600 = 5400.
It is given that the company was able to sell 4/5th of the strips at 25 percent discount and the balance at the printed price of Rs. 250.
Total revenue generated by the firm = $(0.75*250)*(\dfrac{4}{5}*5400)+(250)*(\dfrac{1}{5}*5400)$ = Rs. 1080000
Net revenue after vendor’s discount = 0.7*1080000 = Rs. 756000
We can see that the company invested Rs. 800000 for the drug creation.
Hence, percentage loss incurred by the company = $\dfrac{800000-756000}{800000}*100$ = 5.5%
Therefore, option C is the correct answer.
8) Answer (B)
positive integers n in the range $12 \leq n \leq 40$ such that the product (n -1)*(n – 2)*…*3*2*1 is not divisible by n, implies that n should be a prime no. So there are 7 prime nos. in given range. Hence option B.
9) Answer (A)
a/d = a/b * b/c * c/d = 1/3 * 2 * 1/2 = 1/3
Similarly, b/e and c/f are 3 and 3/8 respectively.
b/e = b/c*c/d*d/e = 3
c/f = c/d*d/e*e/f = 3/8
=> Value of abc/def = 1/3 * 3 * 3/8 = 3/8
10) Answer (C)
$x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$
=> $x = \sqrt{4+\sqrt{4-x}}$
=> $x^2 = 4 + \sqrt{4-x}$
=>$x^4 + 16 – 8x^2 = 4 – x$
=> $x^4 – 8x^2 + x +12 = 0$
On substituting options, we can see that option C satisfies the equation.
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