Competency Focused Questions
Q1: The LCM of two numbers is 48 and one of them is a multiple of other. If these numbers differ by 24, then the sum of the numbers is:
Step 1:Let the smaller number be \(x\) and the larger number be \(kx\), where \(k > 1\) is an integer.
Step 2:Since one number is a multiple of the other, the LCM is the larger number:
\[
kx = 48
\]Step 3:Their difference is given as 24:
\[
kx – x = 24 \\
(k – 1)x = 24
\]Step 4:From \(\,kx = 48\), the larger number is 48.
So, the smaller number is:
\[
48 – 24 = 24
\]Step 5:The sum of the numbers is:
\[
24 + 48 = 72
\]Step 6 (Check):\(\text{LCM}(24,48) = 48 \)
Difference \(= 48 – 24 = 24 \)
Answer: 72 (Note: Correct result is 72, which is not listed in the given options, so likely a misprint in options).
Q2: The ones digit of a number is equal to twice the difference between the digits. The square of this number is divided by half the number. Then 36 is added to quotient and the sum so obtained is divided by 2. The digits of the number so obtained are the same as those in the original number, but in reverse order. The original number is:
Step 1:Let the tens digit = \(x\), and the ones digit = \(y\).
So the original number is:
\[
N = 10x + y
\]Step 2:Condition:
\[
y = 2|x – y|
\]Step 3:Now, square of the number divided by half the number:
\[
\frac{N^2}{\frac{N}{2}} = \frac{N^2}{1}\cdot\frac{2}{N} = 2N
\]
Then add 36:
\[
2N + 36
\]
Now divide by 2:
\[
\frac{2N + 36}{2} = N + 18
\]
So the final number obtained is \(N+18\).
Step 4:Given: The new number is the reverse of the original.
\[
N + 18 = 10y + x
\]
But \(N = 10x + y\). Substituting:
\[
10x + y + 18 = 10y + x \\
9x – 9y + 18 = 0 \\
x – y + 2 = 0 \quad \Rightarrow \quad x = y – 2
\]Step 5:Now use the condition \(y = 2|x-y|\).
Substitute \(x = y – 2\):
\[
|x-y| = |(y-2) – y| = |-2| = 2
\]
So,
\[
y = 2 \times 2 = 4 \\
x = y – 2 = 4 – 2 = 2
\]
Thus the original number is:
\[
N = 10x + y = 10(2) + 4 = 24
\]Step 6 (Verification):Original number = 24.
\[
24^2 = 576, \quad \frac{576}{12} = 48 \\
48 + 36 = 84, \quad \frac{84}{2} = 42
\]
This is the reverse of 24 ✔
Answer: a. 24
Q3: If 5 less than \(\frac{1}{3}\) of a number is same as \(\frac{5}{4}\) less than four times ofthe same number. 6 more than \(\frac{11}{15}\) of the same number is:
Step 1: Let the number be \(x\).
Condition: \(\frac{x}{3} – 5 = 4x – \frac{5}{4}\).
Step 2: Multiply throughout by 12 to remove fractions:
\(12\Big(\frac{x}{3} – 5\Big) = 12\Big(4x – \frac{5}{4}\Big)\)
\(4x – 60 = 48x – 15\)
Step 3: Simplify:
\(4x – 60 = 48x – 15\)
\(-60 + 15 = 48x – 4x\)
\(-45 = 44x\)
\(x = -\frac{45}{44}\)
Step 4: Compute expression: 6 more than \(\frac{11}{15}x\).
= \(\frac{11}{15} \times \frac{-45}{44} + 6\)
= \(-\frac{495}{660} + 6\)
= \(-\frac{3}{4} + 6\)
= \(\frac{24}{4} – \frac{3}{4}\)
= \(\frac{21}{4}\)
Answer: c. \(\frac{21}{4}\)
Q4: If \(0.16\left(5x-2\right)=0.4x+7\), then:
Step 1: Expand the left-hand side:
\(0.16(5x – 2) = 0.16 \times 5x – 0.16 \times 2\)
\(= 0.8x – 0.32\)
Step 2: Form the equation:
\(0.8x – 0.32 = 0.4x + 7\)
Step 3: Bring like terms together:
\(0.8x – 0.4x = 7 + 0.32\)
\(0.4x = 7.32\)
Step 4: Solve for \(x\):
\(x = \frac{7.32}{0.4}\)
\(x = 18.3\)
Answer: d. \(x = 18.3\)
Q5: Rakhi thought of a number, doubled it and added 20 to it. On dividing the resulting number by 25, she get 4. The number is:
Step 1:
Let the number be x.Step 2:
Rakhi doubled the number → \(2x\).
She added 20 → \(2x + 20\).Step 3:
On dividing by 25, the result is 4 →
\[
\frac{2x+20}{25} = 4
\]Step 4:
Multiply both sides by 25:
\[
2x + 20 = 100
\]Step 5:
Subtract 20 from both sides:
\[
2x = 80
\]Step 6:
Divide both sides by 2:
\[
x = 40
\]Answer: b. 40
Q6: In a zoo, total number of monkeys and peacocks is 82. If the total number of legs is 200, then the difference of number of monkeys and peacocks is:
Step 1:Let the number of monkeys = \(x\)
Let the number of peacocks = \(y\)
Step 2:According to the question:
\[
x + y = 82 \tag{1}
\]Step 3:Monkeys have 4 legs, peacocks have 2 legs.
So,
\[
4x + 2y = 200 \tag{2}
\]Step 4:Simplify equation (2):
\[
2x + y = 100 \tag{3}
\]Step 5:From (1): \[
x + y = 82
\]
Subtract (1) from (3):
\[
(2x + y) – (x + y) = 100 – 82 \\
x = 18
\]Step 6:Substitute \(x=18\) into (1):
\[
18 + y = 82 \\
y = 64
\]Step 7:Difference = \(|x – y|\)
\[
|18 – 64| = 46
\]Answer: c. 46
Q7: Tomy, the dog, Tuffy, the puppy, are mother and son respectively. Tuffy is now the same age as that Tomy was four years ago. Four years ago, Tuffy was half his mother’s age. Tomy is now 12 years old. The present age of Tuffy is:
Step 1:Let Tomy’s present age = \(T\) and Tuffy’s present age = \(t\).
Step 2:Given: Tomy is now 12 years old → \(T = 12\).
Step 3:“Tuffy is now the same age as Tomy was four years ago” → \(t = T – 4\).
Substitute \(T=12\):
\[
t = 12 – 4 = 8
\]Step 4 (Verification):Check “Four years ago, Tuffy was half his mother’s age”.
Four years ago: Tuffy’s age = \(t – 4 = 8 – 4 = 4\).
Tomy’s age four years ago = \(T – 4 = 12 – 4 = 8\).
Indeed, \(4 = \frac{1}{2}\times 8\) ✔
Answer: a. 8 years
Q8: Kavita went to market to buy some grocery items. She had some 50 rupee notes and some 100 rupee notes in her purse. The number of 100 rupee notes was twice the number of 50 rupee notes. The total money she had was ₹750. The total number of notes she had in her purse was:
Step 1:Let the number of ₹50 notes = \(x\).
Then the number of ₹100 notes = \(2x\).
Step 2:Total money = Value of ₹50 notes + Value of ₹100 notes
\[
50x + 100(2x) = 750
\]Step 3:Simplify:
\[
50x + 200x = 750 \\
250x = 750 \\
x = \frac{750}{250} = 3
\]Step 4:Number of ₹50 notes = \(x = 3\).
Number of ₹100 notes = \(2x = 6\).
Total number of notes = \(3 + 6 = 9\).
Answer: d. 9
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