CAT Linear Equations Questions PDF [IMPORTANT]

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CAT Linear Equations Questions
CAT Linear Equations Questions

CAT Linear Equations Questions PDF [IMPORTANT]

Linear Equations is a key topic in the CAT Quants section. Make sure you are aware of all the Important Concepts in CAT Quant (Linear Equation). One must not miss out on the questions on Linear Equations in the QA section. Linear Equations fall under the category of Algebra in the CAT Quants; You can also check out these CAT Linear Equation questions from the CAT Previous year papers. This post will look into some important Linear Equation questions for CAT. These are good sources of practice for CAT 2022 preparation. If you want to practice these questions, you can download these Important Linear Equation Questions for CAT (with detailed answers) PDF and the video solutions below, which are completely Free.

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Question 1: A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially?

1) Answer: 62

View Video Solution

Solution:

Let the initial number of chocolates be 64x.

First child gets 32x+1 and 32x-1 are left.

2nd child gets 16x+1/2 and 16x-3/2 are left

3rd child gets 8x+1/4 and 8x-7/4 are left

4th child gets 4x+1/8 and 4x-15/8 are left

5th child gets 2x+1/16 and 2x-31/16 are left.

Given, 2x-31/16=0=> 2x=31/16 => x=31/32.

.’. Initially the Gentleman has 64x i.e. 64*31/32 =62 chocolates.

Question 2: Let A, B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is

a) 5

b) 4

c) 6

d) 7

2) Answer (C)

View Video Solution

Solution:

Given

A+(B+C)/2=5 => 2A+B+C=10….(i)

(A+C)/2 +B=7 => A+2B+C=14…..(ii)

(i)-(ii)=> B-A=4 => B=4+A.

Given, A, B, C are positive integers

If A=1=>B=5 => C=3

If A=2=>B=6 => C=0 but this is invalid as C is positive.

Similarly if A>2 C will be negative and cases are not valid.

Hence, A+B=6.

Question 3: Dick is thrice as old as Tom and Harry is twice as old as Dick. If Dick’s age is 1 year less than the average age of all three, then Harry’s age, in years, is

3) Answer: 18

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Solution:

Let tom’s age = x

=> Dick=3x

=>harry = 6x

Given,

3x+1 = (x+3x+6x)/3

=> x= 3

Hence, Harry’s age = 18 years

Question 4: Let k be a constant. The equations $kx + y = 3$ and $4x + ky = 4$ have a unique solution if and only if

a) $\mid k\mid\neq2$

b) $\mid k\mid=2$

c) $k\neq2$

d) $k=2$

4) Answer (A)

View Video Solution

Solution:

Two linear equations ax+by= c and dx+ ey = f have a unique solution if $\frac{a}{d}\ne\ \frac{b}{e}$

Therefore, $\frac{k}{4}\ne\ \frac{1}{k}$ => $k^2\ne\ 4$

=> k $\ne\ $ |2|

Question 5: In May, John bought the same amount of rice and the same amount of wheat as he had bought in April, but spent ₹ 150 more due to price increase of rice and wheat by 20% and 12%, respectively. If John had spent ₹ 450 on rice in April, then how much did he spend on wheat in May?

a) Rs.560

b) Rs.570

c) Rs.590

d) Rs.580

5) Answer (A)

View Video Solution

Solution:

Let John buy “m” kg of rice and “p” kg of wheat.

Now let the price of rice be “r” in April. Price in May will be “1.2(r)”

Now let the price of wheat be “w” in April . Price in April will be “1.12(w)”.

Now he spent ₹150 more in May , so 0.2(rm)+0.12(wp)=150

Its also given that he had spent ₹450 on rice in April. So (rm)=450

So 0.2(rm)= (0.2)(450)=90 Substituting we get (wp)=60/0.12 or (wp)=500

Amount spent on wheat in May will be 1.12(500)=₹560

Checkout: CAT Free Practice Questions and Videos

Question 6: If x and y are non-negative integers such that $x + 9 = z$, $y + 1 = z$ and $x + y < z + 5$, then the maximum possible value of $2x + y$ equals

6) Answer: 23

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Solution:

We can write x=z-9 and y=z-1 Now we have x+y< z+5

Substituting we get z-9+z-1<z+5 or z<15

Hence the maximum possible value of z is 14

Maximum value of “x” is 5 and maximum value of “y” is 13

Now 2x+y = 10+13=23

Question 7: Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is ₹ 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is

a) 33

b) 27

c) 30

d) 36

7) Answer (A)

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Solution:

Let the number of pencils bought by Aron be “p” and the cost of each pencil be “a”.

Let the number of sharpeners bought Aron be “s” and the cost of each sharpener be “b”.

Now amount spent by Aron will be (pa)+(sb)

Aditya bought (2p) pencils and (s-10) sharpeners. Amount spent will be (2pa)+(s-10)b

Amount spent in both the cases is same

pa + sb = 2pa + (s-10)b or pa=10b

Now its given in the question that cost of sharpener is 2 more than pencil i.e. b=a+2

pa= 10a+20 or a=20/(p-10)

Now the number of pencils has to be minimum, for that we have to find smallest “p” such that both “p” and “a” are integers. The smallest such value is p=11 . Total number of pencils bought will be p+2p=11+22=33

Question 8: A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is

a) 11

b) 13

c) 10

d) 12

8) Answer (B)

View Video Solution

Solution:

Let the cost of an apple, an orange and a mango be a, o, and m respectively.

Then it is given that:

2a+4o+6m = a+4o+8m

or a = 2m.

Also, a+4o+8m = 8o + 7m

10m-7m = 4o

3m = 4o.

We can now express the cost of a basket in terms of mangoes only:

2a+4o+6m = 4m+3m+6m = 13m.

Question 9: If $3x+2\mid y\mid+y=7$ and $x+\mid x \mid+3y=1$ then $x+2y$ is:

a) $-\frac{4}{3}$

b) $\frac{8}{3}$

c) $0$

d) $1$

9) Answer (C)

View Video Solution

Solution:

We need to check for all regions:

x >= 0, y >= 0

x >= 0, y < 0

x < 0, y >= 0

x < 0, y < 0

However, once we find out the answer for any one of the regions, we do not need to calculate for other regions since the options suggest that there will be a single answer.

Let us start with x >= 0, y >= 0,

3x + 3y = 7

2x + 3y = 1

Hence, x = 6 and y = -11/3

Since y > = 0, this is not satisfying the set of rules.

Next, let us test x >= 0, y < 0,

3x – y = 7

2x + 3y = 1

Hence, y = -1

x = 2.

This satisfies both the conditions. Hence, this is the correct point.

WE need the value of x + 2y

x + 2y = 2 + 2(-1) = 2 – 2 = 0.

Question 10: Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?

a) 17

b) 16

c) 18

d) 15

e) 19

10) Answer (C)

View Video Solution

Solution:

If two 50 Misos are used, the 107 can be paid in only 1 way.

If one 50 Miso is used, the number of ways of paying 107 is 6 – zero 10 Miso, one 10 Miso and so on till five 10 Misos.

If no 50 Miso is used, the number of ways of paying 107 is 11 – zero 10 Miso, one 10 Miso and so on till ten 10 Misos.

So, the total number of ways is 18

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