| I. An altitude of a triangle is five-third the length of its corresponding base. If the altitude be increased by 4 cm and the base be decreased by 2 cm the area of the triangle would remain the same. The base and the altitude of the triangle respectively is 12 cm and 20 cm. |
| II. The perimeter of a rectangle is 140 cm. If the length of the rectangle is increased by 2 cm and its breadth decreased by 2 cm the area of the rectangle is increased by 66 sq. cm. The length and breadth of the rectangle respectively is 35 cm and 30 cm. |
| III. The sum of two numbers is 2490. If 6.5% of one number is equal to 8.5% of the other number then one of the numbers will be 1411. |
A)
I
II
III
F
F
F
B)
I
II
III
F
T
T
C)
I
II
III
T
F
F
D)
I
II
III
T
F
T
Correct Answer: D
Solution :
I. Let length of the base of a triangle corresponding to the altitude be \[x\text{ }cm\] Altitude of the triangle \[=\frac{5}{3}xcm\] Area of the triangle\[=\frac{1}{2}\times x\times \frac{5}{3}x=\frac{5}{6}{{x}^{2}}\] According to question, we have \[\frac{1}{2}(x-2)\left( \frac{5}{3}x+4 \right)=\frac{5}{6}{{x}^{2}}\] \[\Rightarrow \left( \frac{1}{2}x-1 \right)\left( \frac{5}{3}x+4 \right)=\frac{5}{6}{{x}^{2}}\] \[\Rightarrow \frac{5}{6}{{x}^{2}}+2x-\frac{5}{3}x-4=\frac{5}{6}{{x}^{2}}\] \[\Rightarrow \frac{5}{6}{{x}^{2}}+2x-\frac{5}{3}x=\frac{5}{6}{{x}^{2}}+4\] \[\Rightarrow \frac{6x-5x}{3}=4\] \[\Rightarrow \frac{x}{3}=4\Rightarrow x=12\] \[\therefore \]Length of base of the triangle = 12 cm Attitude of the triangle \[=\frac{5}{3}\times 12=20cm\] II. Let length of the rectangle be \[x\text{ }cm\] Perimeter of the rectangle = 140 cm \[\Rightarrow \]2(Length + Breadth) = 140 cm \[\Rightarrow \](x+Breadth) \[=\frac{140}{2}=70\] \[\Rightarrow \]Breadlh\[=(70-x)\text{ }cm\] \[\therefore \] Area of the rectangle = Length\[\times \]Breadth \[=x\times (70-x=70x-{{x}^{2}}\] According to question, we have \[(x+2)\text{ }(70-x-2)=70x-{{x}^{2}}+66\] \[\Rightarrow (x+2)\text{ }(68-x)=70x-{{x}^{2}}+66\] \[\Rightarrow 68x-{{x}^{2}}+136-2x=70x-Xs+66\] \[\Rightarrow 66x-{{x}^{2}}-70x+{{x}^{2}}=66-136\] \[\Rightarrow -4x=-70\Rightarrow x=17.5\] \[\therefore \]Length of the rectangle \[=17.5cm\] Breadth of the rectangle\[=(70-17.5)cm\] \[=52.5\text{ }cm\] III. Let the one number be x Then, the other number be 2490 - x 6.5% of the first number \[=x\times \frac{6.5}{100}\] 8.5% of the second number\[=(2490-x)\times \frac{8.5}{100}\] According to question, we have \[\frac{6.5}{100}=(2490-x)\times \frac{8.5}{100}\] \[\Rightarrow \frac{6.5}{100}x+\frac{8.5}{100}x=\frac{21165}{100}\] \[\Rightarrow \frac{15x}{100}=\frac{21165}{100}\Rightarrow x=\frac{21165}{15}\Rightarrow x=1411\]
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