Model Question Paper 2

If 295x703 is divisible by 11, then value of x is:

1. 5

2. 6

3. 7

4. 8

Answer

Divisibility rule of 11 : A number is divisible by 11, if the difference between the sum of the digits at odd places and the sum of the digits at even places (counting from the right) is either 0 or a multiple of 11.

In the number 295x703,

Sum of the digits at odd places = 3 + 7 + 5 + 2 = 17

Sum of the digits at even places = 0 + x + 9 = 9 + x

Difference = 17 - (9 + x) = 8 - x

As the number is divisible by 11, the difference must be 0 or a multiple of 11.

As x is a digit, the only possible value is

⇒ 8 - x = 0

⇒ x = 8

Hence, option 4 is the correct option.

If C.P. of an article is ₹500 and S.P. is ₹600, then his profit % is:

1. 10%

2. 15%

3. 20%

4. 25%

Answer

Given:

C.P. of the article = ₹ 500

S.P. of the article = ₹ 600

As S.P. is greater than C.P., the article is sold at a profit.

Profit = S.P. - C.P.

= ₹ 600 - ₹ 500

= ₹ 100

Profit% = $\dfrac{\text{Profit}}{\text{C.P.}} \times 100$%

$= \dfrac{100}{500} \times 100$%

$= \dfrac{10000}{500}$%

$= 20$%.

Hence, option 3 is the correct option.

Find the values of the letters in the following and give reasons for the steps involved:

$\begin{array}{r r r r} & 4 & B & A \\ +\ & 2 & A & 2 \\ \hline & A & 2 & 9 \\ \hline \end{array}$

Answer

Considering the digits at unit's place :

A + 2 gives 9 at the unit's place.

As A is a digit, A + 2 cannot be 19. So,

⇒ A + 2 = 9

⇒ A = 7 and there is no carry over to the ten's place.

Considering the digits at ten's place :

B + A i.e. B + 7 gives 2 at the ten's place.

As B + 7 cannot be 2, we get

⇒ B + 7 = 12

⇒ B = 5 and 1 is carried over to the hundred's place.

Considering the digits at hundred's place :

4 + 2 + 1 (carry) = 7 = A, which is true as A = 7.

So, the addition is

$\begin{array}{r r r r} & 4 & 5 & 7 \\ +\ & 2 & 7 & 2 \\ \hline & 7 & 2 & 9 \\ \hline \end{array}$

Hence, A = 7 and B = 5.

Krishna bought a pedestal fan for ₹3,068 including 18% GST. Find the price of fan before GST was added.

Answer

Given:

Price of the pedestal fan including GST = ₹ 3,068

Rate of GST = 18%

Let the price of the fan before GST be ₹ $x$.

$\Rightarrow x + \dfrac{18}{100} \times x = 3{,}068 \\[1em] \Rightarrow \dfrac{100x}{100} + \dfrac{18x}{100} = 3{,}068 \\[1em] \Rightarrow \dfrac{118x}{100} = 3{,}068 \\[1em] \Rightarrow x = \dfrac{3{,}068 \times 100}{118} \\[1em] \Rightarrow x = \dfrac{3{,}06{,}800}{118} \\[1em] \Rightarrow x = 2{,}600$

Hence, the price of the fan before GST was added = ₹ 2,600.

Find the amount and compound interest on ₹2,000 for 2 years at 5% per annum, interest payable yearly.

Answer

Given:

Principal (P) = ₹ 2,000

Rate of interest (R) = 5% p.a., interest payable yearly

Number of conversion periods (n) = 2

As we know:

$\Rightarrow \text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 2{,}000\Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 2{,}000\Big(\dfrac{105}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 2{,}000\Big(\dfrac{21}{20}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 2{,}000 \times \dfrac{21}{20} \times \dfrac{21}{20} \\[1em] \Rightarrow \text{A} = ₹ 2{,}205$

C.I. = A - P

= ₹ 2,205 - ₹ 2,000

= ₹ 205

Hence, the amount = ₹ 2,205 and the compound interest = ₹ 205.

If P = {letters of the word HYDERABAD} and Q = {letters of the word ALLAHABAD}, find

(i) P ∪ Q

(ii) P ∩ Q

Also verify that n(P ∪ Q) = n(P) + n(Q) - n(P ∩ Q)

Answer

The distinct letters of the word HYDERABAD are H, Y, D, E, R, A and B.

∴ P = {H, Y, D, E, R, A, B}

The distinct letters of the word ALLAHABAD are A, L, H, B and D.

∴ Q = {A, L, H, B, D}

(i) P ∪ Q contains all the elements which are either in P or in Q or in both.

Hence, P ∪ Q = {H, Y, D, E, R, B, A, L}.

(ii) P ∩ Q contains all the elements which are common to both P and Q.

Hence, P ∩ Q = {A, H, B, D}.

Verification :

⇒ n(P ∪ Q) = 8

⇒ n(P) = 7, n(Q) = 5 and n(P ∩ Q) = 4

⇒ n(P) + n(Q) - n(P ∩ Q) = 7 + 5 - 4 = 8 = n(P ∪ Q)

Hence, n(P ∪ Q) = n(P) + n(Q) - n(P ∩ Q) is verified.

A blanket was marked for ₹500. Shopkeeper allows a discount of 10%. But due to a partial damage in the blanket he has to give an extra discount of 20%. Find the S.P. of the blanket.

Answer

Given:

Marked price of the blanket = ₹ 500

Rate of first discount = 10%

As we know:

$\Rightarrow \text{Discount} = \dfrac{\text{Rate of discount}}{100} \times \text{Price on which discount is given}$

Substituting the values, we get

$\Rightarrow \text{First discount} = ₹ \dfrac{10}{100} \times 500 = ₹ 50$

Price after first discount = ₹ 500 - ₹ 50 = ₹ 450

Rate of second discount = 20%

The second discount is given on the price obtained after the first discount.

$\Rightarrow \text{Second discount} = ₹ \dfrac{20}{100} \times 450 = ₹ 90$

S.P. of the blanket = ₹ 450 - ₹ 90 = ₹ 360

Hence, the S.P. of the blanket = ₹ 360.

In a 3 digit number unit's digit, ten's digit and hundred's digit are in the ratio 3 : 5 : 7. If the difference of original number and number obtained by reversing the digits is 396, find the number.

Answer

Given:

Unit's digit : ten's digit : hundred's digit = 3 : 5 : 7

Let the unit's digit be 3x, the ten's digit be 5x and the hundred's digit be 7x.

Original number = 100 × 7x + 10 × 5x + 3x

= 700x + 50x + 3x = 753x

Number obtained by reversing the digits = 100 × 3x + 10 × 5x + 7x

= 300x + 50x + 7x = 357x

According to given information,

753x - 357x = 396

⇒ 396x = 396

⇒ x = 1

∴ Unit's digit = 3 × 1 = 3, ten's digit = 5 × 1 = 5 and hundred's digit = 7 × 1 = 7

Original number = 753

Hence, the required number = 753.

In an examination, 2500 students appeared out of which 44% are girls. If 50% of the boys and 60% of the girls passed the examination, find the percentage of the students who failed in the examination.

Answer

Given:

Total number of students = 2500

Percentage of girls = 44%

As we know:

$\Rightarrow \text{Number of students} = \dfrac{\text{Percentage of students}}{100} \times \text{Total number of students}$

Substituting the values, we get

$\Rightarrow \text{Number of girls} = \dfrac{44}{100} \times 2500 = 1100$

Number of boys = 2500 - 1100 = 1400

$\Rightarrow \text{Number of boys who passed} = \dfrac{50}{100} \times 1400 = 700\\[1em] \Rightarrow \text{Number of girls who passed} = \dfrac{60}{100} \times 1100 = 660$

Total number of students who passed = 700 + 660 = 1360

Number of students who failed = 2500 - 1360 = 1140

As we know:

$\Rightarrow \text{Percentage of students} = \dfrac{\text{Number of students}}{\text{Total number of students}} \times 100$

Substituting the values, we get

$\Rightarrow \text{Percentage of students who failed} = \dfrac{1140}{2500} \times 100 \\[1em] = \dfrac{114000}{2500} \\[1em] = 45.6$

Hence, 45.6% of the students failed in the examination.

Find the sum invested for $1\dfrac{1}{2}$ years compounded half yearly at the rate of 8% p.a. that will amount to ₹17,576.

Answer

Given:

Amount (A) = ₹ 17,576

Rate of interest = 8% p.a. compounded half yearly

As the interest is compounded half yearly,

Rate per conversion period (R) = $\dfrac{8}{2}$% = 4%

Number of conversion periods (n) = $1\dfrac{1}{2} \times 2$ = 3

As we know:

$\Rightarrow \text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow 17{,}576 = \text{P}\Big(1 + \dfrac{4}{100}\Big)^3 \\[1em] \Rightarrow 17{,}576 = \text{P}\Big(\dfrac{26}{25}\Big)^3 \\[1em] \Rightarrow 17{,}576 = \text{P} \times \dfrac{17{,}576}{15{,}625} \\[1em] \Rightarrow \text{P} = \dfrac{17{,}576 \times 15{,}625}{17{,}576} \\[1em] \Rightarrow \text{P} = 15{,}625$

Hence, the required sum invested = ₹ 15,625.