CAT 2026Papers & G StrategyQuantBreakup G Strategy
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Ages + Ratios
When Rajesh’s age was same as the present age of Garima, the ratio of their ages was 3 : 2. When Garima’s age becomes the same as the present age of Rajesh, the ratio of the ages of Rajesh and Garima will become
1) 2 : 1
2) 3 : 2
3) 4 : 3
4) 5 : 4
Mixtures + Percentages + Ratios
A vessel contained a certain amount of a solution of acid and water. When 2 litres of water was added to it, the new solution had 50% acid concentration. When 15 litres of acid was further added to this new solution, the final solution had 80% acid concentration. The ratio of water and acid in the original solution was
1) 5 : 4
2) 5 : 3
3) 3 : 5
4) 4 : 5
Let:
New water = \( W + 2 \), acid remains \( A \)
New total = \( A + W + 2 \)
Given: 50% acid concentration: \[ \frac{A}{A + W + 2} = \frac{1}{2} \Rightarrow 2A = A + W + 2 \Rightarrow A = W + 2 \tag{1} \]
New acid = \( A + 15 \), water = \( W + 2 \), total = \( A + W + 17 \)
Given: 80% acid concentration: \[ \frac{A + 15}{A + W + 17} = \frac{4}{5} \Rightarrow 5(A + 15) = 4(A + W + 17) \Rightarrow 5A + 75 = 4A + 4W + 68 \Rightarrow A - 4W = -7 \tag{2} \]
From (1): \( A = W + 2 \)
Substituting into (2):
\[
(W + 2) - 4W = -7
\Rightarrow -3W + 2 = -7
\Rightarrow -3W = -9
\Rightarrow W = 3
\]
\[
\Rightarrow A = 3 + 2 = 5
\]
✅ Final Answer: The ratio of water to acid in the original solution is \( \boxed{3 : 5} \)
Logs + Maxima
If \( a, b, c \) are positive real numbers such that:
\[ a > 10 \geq b \geq c \quad \text{and} \quad \frac{\log_8(a + b)}{\log_2 c} + \frac{\log_{27}(a - b)}{\log_3 c} = \frac{2}{3} \]
Then the greatest possible integer value of \( a \) is?
If \( a, b, c \) are positive real numbers such that:
\( a > 10 \geq b \geq c \), and
\[
\frac{\log_8(a + b)}{\log_2 c} + \frac{\log_{27}(a - b)}{\log_3 c} = \frac{2}{3},
\]
then find the greatest possible integer value of \( a \).
Percentages + Averages + Equations
A company has 40 employees whose names are listed in a certain order. In the year 2022, the average bonus of the first 30 employees was Rs. 40,000, of the last 30 employees was Rs. 60,000, and of the first 10 and last 10 employees together was Rs. 50,000. Next year, the average bonus of the first 10 employees increased by 100%, of the last 10 employees increased by 200% and of the remaining employees was unchanged. Then, the average bonus, in rupees, of all the 40 employees together in the year 2023 was
1) 85000 2) 95000 3) 80000 4) 90000
Let the total bonus of the first 10 employees be \( F \), and the last 10 employees be \( L \).
\[ \text{Total bonus} = 30 \times 40{,}000 + L = 12{,}00{,}000 + L \tag{1} \] \[ \text{Total bonus} = 30 \times 60{,}000 + F = 18{,}00{,}000 + F \tag{2} \] Equating (1) and (2): \[ 12{,}00{,}000 + L = 18{,}00{,}000 + F \Rightarrow L - F = 6{,}00{,}000 \tag{3} \]
\[ \frac{F + L}{20} = 50{,}000 \Rightarrow F + L = 10{,}00{,}000 \tag{4} \] Substituting from (3): \( L = F + 6{,}00{,}000 \) into (4): \[ F + (F + 6{,}00{,}000) = 10{,}00{,}000 \Rightarrow 2F = 4{,}00{,}000 \Rightarrow F = 2{,}00{,}000 \] \[ \Rightarrow L = 8{,}00{,}000 \]
\[ \text{Total for first 30} = 30 \times 40{,}000 = 12{,}00{,}000 \Rightarrow \text{Middle 20} = 12{,}00{,}000 - F = 10{,}00{,}000 \]
\[ \text{Total bonus} = 4{,}00{,}000 + 10{,}00{,}000 + 24{,}00{,}000 = 38{,}00{,}000 \] \[ \text{Average bonus} = \frac{38{,}00{,}000}{40} = \boxed{95{,}000} \]
✅ Final Answer: Rs. 95,000
Coordinate + InCircle
The coordinates of the three vertices of a triangle are: (1, 2), (7, 2), and (1, 10). Then the radius of the in circle of the triangle is ________
The coordinates of the vertices of a triangle are: \( (1, 2), (7, 2), (1, 10) \). Find the radius of the incircle of the triangle.
When plotted, the triangle clearly forms a right-angled triangle with the right angle at point \( (1, 2) \). This is because two sides are aligned along the x-axis and y-axis.
Let side \( a \) be the vertical leg: \[ a = 10 - 2 = 8 \] Let side \( b \) be the horizontal leg: \[ b = 7 - 1 = 6 \]
Using the Pythagorean Theorem: \[ \text{Hypotenuse } h = \sqrt{a^2 + b^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \]
The inradius of a right triangle is given by: \[ r = \frac{a + b - h}{2} \] Substituting the values: \[ r = \frac{8 + 6 - 10}{2} = \frac{4}{2} = 2 \]
The radius of the incircle is: \( \boxed{2} \)
Profit Loss + Percentages
Bina incurs 19% loss when she sells a product at Rs. 4,860 to Shyam, who in turn sells this product to Hari. If Bina would have sold this product to Shyam at the purchase price of Hari, she would have obtained 17% profit. Then, the profit, in rupees, made by Shyam is _______
Bina incurs a 19% loss when she sells a product at Rs. 4,860 to Shyam, who in turn sells this product to Hari. If Bina would have sold this product to Shyam at the purchase price of Hari, she would have obtained a 17% profit. Then, the profit, in rupees, made by Shyam is:
Let the cost price of the product for Bina be \( x \). Given she sold it to Shyam at a 19% loss: \[ 4860 = x \times (1 - 0.19) = 0.81x \] Solving: \[ x = \frac{4860}{0.81} = 6000 \] So, Bina's cost price was Rs. 6000.
If Bina had sold it at Hari's purchase price and made a 17% profit: \[ \text{Hari's Purchase Price} = 6000 \times 1.17 = 7020 \]
Shyam bought it at Rs. 4860 and sold it at Rs. 7020, so his profit was: \[ 7020 - 4860 = \boxed{2160} \]
\( \boxed{\text{Rs. } 2160} \)
Algebra + APGP + Fractions
Find the sum of the series
\[ \frac{1}{5} \left( \frac{1}{5} - \frac{1}{7} \right) + \left( \frac{1}{5} \right)^2 \left( \left( \frac{1}{5} \right)^2 - \left( \frac{1}{7} \right)^2 \right) + \left( \frac{1}{5} \right)^3 \left( \left( \frac{1}{5} \right)^3 - \left( \frac{1}{7} \right)^3 \right) + \cdots \]
Given Series:
\[ \frac{1}{5} \left( \frac{1}{5} - \frac{1}{7} \right) + \left( \frac{1}{5} \right)^2 \left( \left( \frac{1}{5} \right)^2 - \left( \frac{1}{7} \right)^2 \right) + \left( \frac{1}{5} \right)^3 \left( \left( \frac{1}{5} \right)^3 - \left( \frac{1}{7} \right)^3 \right) + \cdots \]
\[ \sum_{n=1}^{\infty} \left( \frac{1}{5} \right)^n \left[ \left( \frac{1}{5} \right)^n - \left( \frac{1}{7} \right)^n \right] = \sum_{n=1}^{\infty} \left[ \left( \frac{1}{25} \right)^n - \left( \frac{1}{35} \right)^n \right] \] This is a difference of two geometric series.
Using the formula \( \sum_{n=1}^{\infty} r^n = \frac{r}{1 - r} \), we get: \[ \sum_{n=1}^{\infty} \left( \frac{1}{25} \right)^n = \frac{1/25}{1 - 1/25} = \frac{1}{24} \] \[ \sum_{n=1}^{\infty} \left( \frac{1}{35} \right)^n = \frac{1/35}{1 - 1/35} = \frac{1}{34} \]
\[ \sum = \frac{1}{24} - \frac{1}{34} = \frac{34 - 24}{24 \cdot 34} = \frac{10}{816} = \frac{5}{408} \]
\[ \boxed{\frac{5}{408}} \]
A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is ______ TITA
A fruit seller has a stock of mangoes, bananas and apples, with at least one fruit of each type. Initially, mangoes make up 40% of the stock. He sells:
Let total fruits be \( x \). Then:
Fruits sold:
Total fruits sold = 50% of total stock: \[ 0.2x + 96 + 0.24x - 0.4b = 0.5x \Rightarrow 0.44x - 0.4b + 96 = 0.5x \Rightarrow 96 - 0.4b = 0.06x \]
Multiply both sides by 100: \[ 9600 - 40b = 6x \Rightarrow x = \frac{9600 - 40b}{6} \Rightarrow x = \frac{20(240 - b)}{3} \] For \( x \) to be a whole number, \( (240 - b) \) must be divisible by 3. So try smallest values of \( b \) such that: \[ b \equiv 0 \pmod{3} \]
\[ x = \frac{20(240 - 99)}{3} = \frac{20 \times 141}{3} = 940 \] Check:
\( \boxed{940} \)
Find the value of x? \[ \left( x + 6\sqrt{2} \right)^{\frac{1}{2}} - \left( x - 6\sqrt{2} \right)^{\frac{1}{2}} = 2\sqrt{2}, \]
We are given the equation:
\[ \sqrt{x + 6\sqrt{2}} - \sqrt{x - 6\sqrt{2}} = 2\sqrt{2} \]
Using the identity \( (a - b)^2 = a^2 + b^2 - 2ab \):
\[ (\sqrt{x + 6\sqrt{2}} - \sqrt{x - 6\sqrt{2}})^2 = (2\sqrt{2})^2 \] \[ = (x + 6\sqrt{2}) + (x - 6\sqrt{2}) - 2\sqrt{(x + 6\sqrt{2})(x - 6\sqrt{2})} \] \[ = 2x - 2\sqrt{x^2 - 72} = 8 \]
Divide both sides by 2:
\[ x - \sqrt{x^2 - 72} = 4 \]
\[ \sqrt{x^2 - 72} = x - 4 \] \[ \Rightarrow x^2 - 72 = (x - 4)^2 = x^2 - 8x + 16 \] \[ \Rightarrow -72 = -8x + 16 \Rightarrow 8x = 88 \Rightarrow x = \boxed{11} \]
\[ \boxed{x = 11} \]
Anil invests Rs 22000 for 6 years in a scheme with 4% interest per annum, compounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives at the end of 5 years, for one year at 10% simple interest. If the amounts received by both at the end of 6 years are equal, then the initial investment, in rupees, made by Sunil is
Anil invests ₹22000 for 6 years in a scheme offering 4% interest per annum, compounded half-yearly. Sunil invests a certain amount in the same scheme for 5 years and then reinvests the entire amount for 1 year at 10% simple interest. If both receive the same amount at the end of 6 years, find Sunil's initial investment.
Anil’s Investment:
Compound interest formula:
\[
A = P \left(1 + \frac{r}{100}\right)^n
\]
- Principal \( P = 22000 \)
- Rate per half-year = \( \frac{4}{2} = 2\% \)
- Time = \( 6 \times 2 = 12 \) half-years
\[
A_{\text{Anil}} = 22000 \times (1.02)^{12} \approx 22000 \times 1.26824 = 27901.28
\]
Sunil’s Investment:
Let the initial investment be \( P \).
For 5 years = 10 half-years:
\[
A_{\text{5 years}} = P \times (1.02)^{10} \approx P \times 1.219
\]
Reinvested for 1 year at 10% simple interest:
\[
A_{\text{Sunil}} = A_{\text{5 years}} \times \left(1 + \frac{10}{100}\right) = P \times 1.219 \times 1.10 = P \times 1.3409
\]
Equating to Anil’s final amount: \[ P \times 1.3409 = 27901.28 \Rightarrow P = \frac{27901.28}{1.3409} \approx \boxed{20812} \]
\( \boxed{₹20812} \)