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1
The monthly incomes of A and B are in ratio 3 : 5 and the ratio of their saving is 2 : 3. If the income of B is equal to three times the saving of A, then what is the ratio of the expenditures of A and B?
Discuss
Answer & Solution
Answer: Option B
Solution:
\[\begin{array}{*{20}{c}} {}&{\text{A}}&{}&{\text{B}} \\ {{\text{Income}} \to }&{3x}&:&{5x} \\ {{\text{Saving}} \to }&{2y}&:&{3y} \end{array}\]
$$\eqalign{ & 5x = 3 \times 2y \cr & \frac{x}{y} = \frac{6}{5} \cr} $$
\[\begin{array}{*{20}{c}} {}&{\,{\text{A}}\,\,\,\,\,{\text{B}}\,} \\ {{\text{Income}} \to }&{18\,\,\,\,30} \\ {{\text{Saving}} \to }&{10\,\,\,\,15} \\ {{\text{Expenditure}} \to }&{\overline {\,\,\,8\,:\,15\,\,} } \end{array}\]
2
Rs. 6,300 is divided between X, Y, Z such that X : Y = 7 : 5 and Y : Z = 4 : 3. Find the share of Y.
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & x:y:z \cr & 7:5 \cr & \,\,\,\,\,\,\,\,4:3 \cr & \overline {\,28:20:15\,} \cr & \Rightarrow 63\,{\text{units}} \to 6300 \cr & 1\,{\text{unit}} \to 100 \cr & 20{\text{ units}} \to 20 \times 100 = 2000 \cr} $$
3
A certain sum is divided between A, B, C and D such that the ratio of the shares of A and B is 1 : 3, that of B and C is 2 : 5, and that of C and D 2 : 3. If the difference the shares of A and C is Rs. 3,510, then the share of D is:
Discuss
Answer & Solution
Answer: Option C
Solution:
According to the question
\[\begin{array}{*{20}{c}} {\text{A}}&:&{\text{B}}&:&{\text{C}}&:&{\text{D}} \\ 1&:&3&{}&{}&{}&{} \\ {}&{}&2&:&5&{}&{} \\ {}&{}&{}&{}&2&:&3 \end{array}\]
The ratio of the A : B : C : D is
⇒ (1 × 2 × 2) : (3 × 2 × 2) : (3 × 5 × 2) : (3 × 5 × 3)
⇒ 4 : 12 : 30 : 45
The difference between the share of A and C is
⇒ (30 - 4) units
⇒ 26 units
Now,
⇒ 26 units = 3510
⇒ 1 unit = $$\frac{{3510}}{{26}}$$
⇒ 1 unit = Rs. 135
The share of D is
⇒ 45 units
⇒ 45 × 135
⇒ 6075
The share of D is Rs. 6075
4
The income of A is $$\frac{2}{3}$$ of B's income and the expenditure of A is $$\frac{3}{4}$$ of B's expenditure. If $$\frac{1}{3}$$ of the income of B is equal to the expenditure of A, then the ratio of the savings of A to those of B is:
Discuss
Answer & Solution
Answer: Option A
Solution:
Income of A is equal to $$\frac{2}{3}$$ of income of B.
$$\eqalign{ & A = B \times \frac{2}{3} \cr & \frac{A}{B} = \frac{{2x}}{{3x}} \cr} $$
A's expenditure is equal to $$\frac{3}{4}$$ B's expenditure
$$\eqalign{ & A = B \times \frac{3}{4} \cr & \frac{A}{B} = \frac{{3y}}{{4y}} \cr} $$
A's income is equal to $$\frac{1}{3}$$ B's income
3x × $$\frac{1}{3}$$ = 3y
x = 3y
Saving ratio of A and B
= (2x - 3y) : (3x - 4y)   [∵ x = 3y]
= (6y - 3y) : (9y - 4y)
= 3y : 5y
= 3 : 5

Alternate solution
\[\begin{array}{*{20}{c}} {}&{{\text{A}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{B}}} \\ {{\text{Income}} \to }&{\,\,\,\,\,\,\,{2_{ \times 3 = 6}}\,\,\,{3_{ \times 3 = 9}}} \\ {{\text{Expenditure}} \to }&{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,4} \\ {{\text{Saving}} \to }&{\overline {\underline {\,\,3\,\,\,\,\,:\,\,\,\,\,5\,\,} } } \end{array}\]
5
The ratio of the monthly incomes of X and Y is 5 : 4 and that of their monthly expenditures is 9 : 7. If the income of Y is equal to the expenditure of X, then what is the ratio of the saving of X and Y?
Discuss
Answer & Solution
Answer: Option A
Solution:
\[\begin{array}{*{20}{c}} {}&{{\text{X}}\,\,\,\,\,:\,\,\,\,\,{\text{Y}}} \\ {{\text{Income}} \to }&{{5_{ \times 9}}\,\,\,:\,\,\,{4_{ \times 9}}} \\ {{\text{Expenditure}} \to }&{{9_{ \times 4}}\,\,\,:\,\,\,{7_{ \times 4}}} \\ {{\text{Saving}} \to }&{\overline {\underline {\,\,9\,\,\,\,\,\,:\,\,\,\,\,\,8\,\,} } } \end{array}\]
Y's income = X's Expenditure
Saving = 9 : 8
6
A sum of Rs. 1250 has to distributed among A, B, C and D. Total share of B and D is equal to (14/11) of total share of A and C. Share of D is half of share of A. Share of C is 1.2 of share of A. What are the shares of A, B, C and D respectively?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{A}} + {\text{B}} + {\text{C}} + {\text{D}} = 1250.......\left( 1 \right) \cr & \frac{{{\text{B}} + {\text{D}}}}{{{\text{A}} + {\text{C}}}} = \frac{{14}}{{11}}.......\left( 2 \right) \cr & \frac{{\text{D}}}{{\text{A}}} = \frac{1}{2} \cr & \frac{{\text{C}}}{{\text{A}}} = \frac{{1.2}}{1} = \frac{6}{5} \cr & {\text{From equation }}\left( 1 \right) \cr & 14 + 11 = 1250 \cr & 25{\text{ units}} = {\text{1250}} \cr & {\text{1 unit}} = 50 \cr & {\text{C's share 6 units}} \to {\text{300}} \cr & {\text{A's share 5 units}} \to {\text{250}} \cr} $$
\[ \Rightarrow \frac{{\text{D}}}{{\text{A}}} = \frac{1}{2}\begin{array}{*{20}{c}} { \to 125} \\ { \to 250} \end{array}\]
$$\eqalign{ & {\text{From equation }}\left( 1 \right) \cr & {\text{A}} + {\text{B}} + {\text{C}} + {\text{D}} = 1250 \cr & 250 + {\text{B}} + 300 + 125 = 1250 \cr & {\text{B}} = 575 \cr} $$
7
The ratio of boys and girls in a school is 27 : 23. If the difference between the number of boys and girls is 200, then find the number of boys.
Discuss
Answer & Solution
Answer: Option A
Solution:
Let the number of boys and girls be 27x and 23x respectively
According to condition
27x - 23x = 200
⇒ x = 50
The number of boys = 27 × 50
⇒ 1350
∴ The number of boys is 1350
8
Alloy A contains metals x and y only in the ratio 5 : 2 and alloy B contains these metals in the ratio 3 : 4. Alloy C is prepared by mixing A and B in the ratio 4 : 5. The percentage of x in alloy C is:
Discuss
Answer & Solution
Answer: Option B
Solution:
\[\begin{array}{*{20}{c}} {}&x&:&y&{} \\ {A \to }&{{5_{ \times 4}}}&:&{{2_{ \times 4}}}&{ = {7_{ \times 4}}} \\ {B \to }&{{3_{ \times 5}}}&:&{{4_{ \times 5}}}&{ = {7_{ \times 5}}} \end{array}\]
$$\eqalign{ & C \to A + B \to 35 + 28 = 63 \cr & \% \,{\text{of }}x{\text{ in aloy C}} = \frac{{35}}{{63}} \times 100 = 55\frac{5}{9}\% \cr} $$
9
If $$a:b:c = \frac{1}{4}:\frac{1}{3}:\frac{1}{2},$$    then $$\frac{a}{b}:\frac{b}{c}:\frac{c}{a} = ?$$
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & a:b:c = \frac{1}{4}:\frac{1}{3}:\frac{1}{2} \cr & a:b:c = 3:4:6 \cr & \frac{a}{b}:\frac{b}{c}:\frac{c}{a} = \frac{3}{4}:\frac{4}{6}:\frac{6}{3} = 9:8:24 \cr} $$
10
The sum of three numbers is 280. If the ratio between the first and second numbers is 2 : 3 and the ratio between second and third numbers is 4 : 5, then find the second number.
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & \,\,\,{{\text{I}}^{{\text{st}}}}\,\,\,{\text{I}}{{\text{I}}^{{\text{nd}}}}\,\,\,{\text{II}}{{\text{I}}^{{\text{rd}}}} \cr & \,\,2\,\,:\,\,3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,:\,\,5 \cr & \overline {\,8\,:\,12\,:\,15\,} \cr & 35{\text{ unit}} \to {\text{280}} \cr & {\text{1 unit}} \to 8 \cr & 12\,{\text{unit}} \to 12 \times 8 = 96 \cr} $$