CAT 2021 QA Slot 1 | Previous Year CAT Paper

Answer: 35

Video Explanation

Explanation :

Let the number of patients in hospital A = x
∴ The number of patients in hospital B = x + 21

⇒ $\frac{200}{x}$ = $\frac{152}{x+21}$ + 3

⇒ 200x + 4200 = 152x + 3x² + 63x

⇒ 3x² + 15x - 4200 = 0

⇒ x² + 5x - 1400 = 0

⇒ x = 35 or -40 (not possible)

∴ Number of patients in hospital A = 35

Hence, 35.

Answer: 1000

Video Explanation

Explanation :

Let the total number of pens Amal bought = x

Also, let the wage of the employee = w

∴ Amal’s total cost price = 8x + w

Total selling price of the first 100 pen = 100 × 12 = 1200

Case 1: The remaining pen when each is sold at Rs. 11
Total selling price of the remaining pen = (x - 100) × 11

⇒ 1200 + 11(x - 100) = 8x + w + 300
⇒ 3x - 200 = w   …(1)

Case 2: The remaining pen when each is sold at Rs. 9
Total selling price of the remaining pen = (x - 100) × 9

⇒ 1200 + 9(x - 100) = 8x + w - 300
⇒ x + 600 = w   …(2)

Solving (1) and (2) we get

x = 400 and w = 1000.

Hence, 1000.

Answer: Option C

Video Explanation

Explanation :

Let the earnings of Neeta, Geeta and Sita be ‘n’, ‘g’ and ‘s’ respectively.

The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days
⇒ 6s = n + g   …(1)

The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days
⇒ 2g = s + n
⇒ 2(6s – n) = s + n [From (1)]
⇒ 11s = 3n
⇒ n = 11s/3   …(2)

Substituting n = 11s/3 in (1)

⇒ 6s = 11s/3 + g
⇒ 7s/3 = g   …(3)

From (2) and (3)
11s/3 > 7s/3 > s

∴ n > g > s

∴ The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is = 11s/3 : s = 11 : 3

Hence, option (c).

Answer: Option D

Video Explanation

Explanation :

Here the critical points are 20, 60 and 100.

Case 1: n ≥ 100
⇒ n - 60 < n - 100
⇒ -60 < -100
This can never be true.
This case is rejected.

Case 2: 60 ≤ n < 100
⇒ n - 60 < - n + 100
⇒ 2n < 160
⇒ n < 80

Also,
- n + 100 < n – 20
⇒ 2n > 120
⇒ n > 60

∴ Possible integral values of n are 61, 62, 63, …, 79 i.e., 19 values

Case 3: n < 60
⇒ - n + 100 < - n + 20
⇒ 100 < 20
This can never be true.
This case is rejected.

∴ n can take 19 integral values.

Hence, option (d).

Answer: Option C

Video Explanation

Explanation :

Given, |x² – 4x - 13| = r
∴ x² – 4x - 13 = ± r

We have two quadratic equations here but only three distinct roots it means one of the quadratic equations will have equal roots.

Case 1: x² – 4x - 13 = r has equal roots, i.e., Discriminant = 0
⇒ x² – 4x – 13 - r = 0 had D = 0

⇒ D = 16 – 4(-13 - r) = 0
⇒ 16 + 52 + 4r = 0
⇒ r = - 17 

Case 2: x² – 4x - 13 = - r has equal roots, i.e., Discriminant = 0
⇒ x² – 4x – 13 + r = 0 had D = 0

⇒ D = 16 – 4(-13 + r) = 0
⇒ 16 + 52 - 4r = 0
⇒ r = 17

Hence, option (c).

Answer: Option C

Video Explanation

Explanation :

Let the length of the smaller train be L meters.
Also, let the speeds of faster and slower trains be ‘f’ and ‘s’ m/s.

Given, f - s = 6 × 5/18 = 5/3 m/s
⇒ f = s + 5/3   …(1)

Faster train crosses a lamp post in 12 seconds.
∴ 160/f = 12
⇒ f = 40/3

∴ s = f – 5/3 = 35/3

The two trains cross each other in 14 seconds.
∴ (160 + L)/(f + s) = 14
⇒ L = 14(f + s) – 160
⇒ L = 14 × 25  – 160
⇒ L = 350 – 160 = 190

Hence, option (c).

Concept: Trains

Answer: Option D

Video Explanation

Explanation :

Let the total work to be done = LCM (15, 12, 20) = 60 units.
Efficiency of Anu = 60/15 = 4 units/day
Efficiency of Vinu = 60/12 = 5 units/day
Efficiency of Manu = 60/20 = 3 units/day

Let us calculate the work done every 2 days.
Vinu works on both the days, hence work done by Vinu = 2 × 5 = 10 units
Anu works on alternate days, hence work done by Anu = 4 units
Manu works on alternate days, hence work done by Manu = 3 units

∴ Total work done in 2 days = 10 + 4 + 3 = 17 units.

Hence total work done in 6 days = 3 × 17 = 51 units.

On 7th day Vinu and Anu will work, hence work done = 5 + 4 = 9 units.

∴ Work done in total 7 days = 51 + 9 = 60 units [Work gets completed now]

∴ The number of days needed to complete the work is 7 days.

Hence, option (d).

Answer: 47

Video Explanation

Explanation :

Out of the seven integers given, 3 and 5 should always be selected.

Case 1: Exactly 1 of 7 or 8 is selected.
Exactly one of 7 or 8 can be selected in 2 ways.
Now we have already selected three integers.

For each of the remaining 4 integers, it may be selected or it may not be selected.
∴ Number of ways of selection for remaining integers = 2 × 2 × 2 × 2 = 16

∴ Total number of ways = 2 × 16 = 32

Case 2: None of 7 and 8 is selected.
So far we have selected only two integer.

For each of the remaining 4 integers, it may be selected or it may not be selected.
∴ Number of ways of selection for remaining integers = 2 × 2 × 2 × 2 = 16
But this includes one way when no integer is selected.
∴ Number of ways of selecting at least one integer = 16 – 1 = 15

∴ Total number of ways = 32 + 15 = 47.

Hence, 47.

Answer: Option D

Video Explanation

Explanation :

Area of the equilateral triangle = √3/4 × (12)²

Let the length of each side of the regular hexagon be ‘a’.
Area of a regular hexagon with side ‘a’ = 6 × √3/4 × (a)²

Now, 6 × √3/4 × (a)² = √3/4 × (12)²

⇒ a² = 24

⇒ a = 2√6

Hence, option (d).

Answer: 99

Video Explanation

Explanation :

Given,
5 – ${\log }_{10}\left(\sqrt{1+x}\right)$ + 4${\log }_{10}\left(\sqrt{1-x}\right)$ = ${\log }_{10}\left(\frac{1}{\sqrt{1-x^2}}\right)$

⇒  5 – ${\log }_{10}{\left(1+x\right)}^{1/2}$ + ${\log }_{10}{\left(1-x\right)}^{4/2}$ = ${\log }_{10}{\left(1-x^2\right)}^{-1/2}$

⇒  5 = ${\log }_{10}{\left(1+x\right)}^{1/2}$ - ${\log }_{10}{\left(1-x\right)}^2$ - ${\log }_{10}{\left(1-x^2\right)}^{1/2}$

⇒  5 = ${\log }_{10}\left[\frac{{\left(1+x\right)}^{1/2}}{{(1-x)}^2\times {\left\{\right(1+x\left)\right(1-x\left)\right\}}^{1/2}}\right]$

⇒  5 = ${\log }_{10}{(1-x)}^{-5/2}$

⇒ 5 = -5/2 × log₁₀(1 - x)

⇒ -2 = log₁₀(1 - x)

⇒ 1 – x = 10^-2 = 1/100

⇒ 100 – 100x = 1

⇒ 100x = 99

Hence, 99.

Answer: Option A

Video Explanation

Explanation :

Let the cost of each apple be ‘a’, each orange be ‘o’ and each mango be ‘m’.

∴ 2a + 4o + 6m = a + 4o + 8m = 8o + 7m

⇒ 2a + 4o + 6m = 8o + 7m
⇒ 4o + m = 2a   …(1)

Also, a + 4o + 8m = 8o + 7m
⇒ 4o – m = a   …(2)

Solving (1) and (2) we get
8o = 3a and m = a/2

∴ Value of each basked in terms of mangoes
= 2a + 4o + 6m
= 4m + 3a/2 + 6m
= 10m + 3/2 × 2m
= 10m + 3m
= 13m

∴ Value of each basket is same as value of 13 mangoes.

Hence, option (a).

Answer: 200

Video Explanation

Explanation :

Let x cc is removed from first bottle and replaced with solution from second bottle.

Using alligation rule

⇒ (800-x)/x = 4/12 = 1/3

⇒ 2400 – 3x = x

⇒ x = 600 cc.

∴ Solution left in second bottle = 800 – 600 = 200 cc.

Hence, 200.

Answer: Option C

Video Explanation

Explanation :

Let the amounts spent by the family each month be LCM (10, 20, 25, 50) = Rs. 100 for the first 3 months and then Rs. 50 for the next two months.

Amount of onion bought during month 1 = 100/10 = 10 kgs
Amount of onion bought during month 2 = 100/20 = 5 kgs
Amount of onion bought during month 3 = 100/25 = 4 kgs
Amount of onion bought during month 4 = 50/25 = 2 kgs
Amount of onion bought during month 5 = 50/50 = 1 kgs

∴ Total amount of onion bought = 10 + 5 + 4 + 2 + 1 = 22 kgs
Total amount spend on onions = 100 + 100 + 100 + 50 + 50 = Rs. 400.

∴ Average expense for onion per kg for these 5 months = 400/22 = 18.18 ≈ Rs. 18/kg.

Hence, option (c). 

Answer: Option C

Video Explanation

Explanation :

Side of the regular hexagon = 2 cm.
Consider the figure below.

​​​​​​​

Consider the isosceles ∆ATF

TU is the altitude from T to AF. 
We know, in a regular hexagon the distance between any two parallel sides = √3 × side.
∴ TU = √3 × 2 = 2√3 cm.

Since ATF is an isosceles triangle, U will be the mid-point of AF
∴ AU = 1 cm.

In ∆ATU
AT² = TU² + AU²
⇒ AT² = (2√3)² + 1²
⇒ AT² = 13
⇒ AT = √13

Hence, option (c).

Answer: 120

Video Explanation

Explanation :

Here, BM = BN = 4 (Radius of the circle)

BC = 10 cm, hence, MC = CO = 10 – 4 = 6 cm.

Let AN = x, hence, AO = x

In right ∆ABC,
AB² + BC² = AC²

⇒ (x + 4)² + 10² = (x + 6)²

⇒ x² + 16 + 8x + 100 = x²+ 12x + 36

⇒ 4x = 80

⇒ x = 20

∴ Area of triangle ABC = ½ × AB × BC = ½ × (20 + 4) × 10 = 120 sq. inches.

Hence, 120.

Answer: Option C

Video Explanation

Explanation :

Given, f(x) = $\frac{x^2+2x-15}{x^2-7x-18}$ < 0

⇒ $\frac{(x+5)(x-3)}{(x-9)(x+2)}$  < 0

Here, the critical points are -5, -2, 3 and 9.

∴ f(x) will be positive when -5 < x < -2 or 3 < x < 9.

Hence, option (c).

Answer: 32

Video Explanation

Explanation :

Let the total work to be done = LCM (12, 16, 24) = 48 units.

Let the efficiency of Amar, Akbar and Anthony be ‘x’, ‘y’ and ‘z’ units/month

⇒ x + y = 48/12 = 4   …(2)
⇒ y + z = 48/16 = 3   …(1)
⇒ z + x = 48/24 = 2   …(3)

Adding all these equations
⇒ x + y + z = 9/2 = 4.5   …(4)

Solving these four equations we get,
x = 1.5, y = 2.5 and z = 0.5

∴ The person who is neither slowest not fastest is Amar with efficiency of 1.5 units/month.

∴ Time required by Amar to complete the task alone = 48/1.5 = 32 months.

Hence, 32.

Answer: Option A

Video Explanation

Explanation :

Let the weight and price of small box is ‘w’ and ‘p’ respectively.
∴ Selling price per gram for small box = p/w

Now, selling price per gram for large box = 0.88p/w

Also, selling price for the whole box = 2p.

⇒ Weight of the large box = $\frac{2p}{0.88p/w}$ = $\frac{2w}{0.88}$ = $\frac{25w}{11}$

∴ Required percentage = $\frac{{\displaystyle \frac{25}{11}}w-w}{w}$ × 100 = 14/11 × 100 ≈ 127%

Hence, option (a).

Answer: 70

Video Explanation

Explanation :

Let the required three-digit number be ‘abc’.

According to the question:

‘abc’ + 198 = ‘cba’

⇒ 100a + 10b + c + 198 = 100c + 10b + a

⇒ 99c – 99a = 198

⇒ c – a = 2

∴ Value of (c, a) can be (9, 7) or (8, 6) or (7, 5) or (6, 4) or (5, 3) or (4, 2) or (3, 1) i.e., total 7 values.

For each of these 7 values of c and a, b can taken any of the 10 values from 0 to 9.

∴ Total such possible numbers = 7 × 10 = 70.

Hence, 70.

Answer: Option C

Video Explanation

Explanation :

Let the interest accrued during first year = I and rate of interest be r%

We know in case of compound interest, interest for each year increases by r% every year.

∴ Interest for 2^nd year = I × $\left(1+\frac{r}{100}\right)$ = 806.25   …(1)
Interest for 3^rd year = I × ${\left(1+\frac{r}{100}\right)}^2$ = 866.72   …(2)

(2) ÷ (1)

⇒ $\left(1+\frac{r}{100}\right)$ = $\frac{866.72}{806.25}$ = 1 - $\frac{60.47}{806.25}$

⇒ r = 7.5%

∴ Interest for 4^th year = Interest for 3^rd year × (1+r/100)

= 866.72 × $\left(1+\frac{7.5}{100}\right)$ = 931.72

Hence, option (c).

Answer: Option D

Video Explanation

Explanation :

Given, x_n+2 = $\frac{1+x_{n+1}}{x_n}$
x₀ = 1, x₁ = 2

Substituting n = 0 we get

x₂ = $\frac{1+2}{1}$ = 3

Similarly, 
x₃ = 2
x₄ = 1

x₅ = 1
x₆ = 2
x₇ = 3
x₈ = 2
x₉ = 1
… and so on.

Here, we can see that the value of xn repeats after every 5 terms.

∴ x₂₀₂₁ = x₂₀₂₀₊₁ = x₁ = 2

Hence, option (d).

Answer: Option B

Video Explanation

Explanation :

Here, 
1^st group has 1 integer
2^nd group has 3 integers
3^rd group has 5 integers
∴ nth groups will have (2n – 1) integers.

Total inters used till
1^st group = 1 = 1²
2^nd group = 1 + 3 = 2²
3^rd group = 1 + 3 + 5 = 9 = 3²

∴ Total integers used till n^th group = n²

∴ Sum of integers in 15^th group = Sum of all integers used till 15^th group - Sum of all integers used till14^th group

= (1 + 2 + 3 + … + 15²) - (1 + 2 + 3 + … + 14²)

= $\frac{225\times 226}{2}$ - $\frac{196\times 197}{2}$

= 25425 – 19306 = 6119

Hence, option (b).