Tick (✓) the correct answer
Question: Tick (✓) the correct answer If 7x8 is exactly divisible by 9, then the least value ofxis (a) 0 (b) 2 (c) 3(d) 5 Solution: (c) 3 If a number is exactly divisible by 9, the sum of the digits must also be divisible by 9. $7+x+8=15+x$ 18 is divisible by 9 . $\therefore 15+x=18 \Rightarrow x=3$...
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Question: Tick (✓) the correct answer If 64y8 is exactly divisible by 3, then the least value ofyis (a) 0 (b) 1 (c) 2 (d) 3 Solution: (a) 0 If a number is divisible by 3, then the sum of the digits is also divisible by 3. $6+4+y+8=18+y$ This is divisible by 3 as $y$ is equal to 0 ....
Read More →Find the number of revolutions made
Question: Find the number of revolutions made by a circular wheel of area 1.54 m2in rolling a distance of 176 m. Solution: Let the number of revolutions made by a circular wheel be n and the radius of circular wheel be r. Given that, $\quad$ area of circular wheel $=1.54 \mathrm{~m}^{2}$ $\Rightarrow \quad \pi r^{2}=1.54$$\left[\because\right.$ area of circular $\left.=\pi r^{2}\right]$ $\Rightarrow$ $r^{2}=\frac{1.54}{22} \times 7 \Rightarrow r^{2}=0.49$ $\therefore$ $r=0.7 \mathrm{~m}$ So, the...
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer If 64y8 is exactly divisible by 3, then the least value ofyis (a) 0 (b) 1 (c) 2 (d) 3 Solution: (a) 0 If a number is divisible by 3, then the sum of the digits is also divisible by 3. $6+4+y+8=18+y$ This is divisible by 3 as $y$ is equal to 0 ....
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer If 5x6 is exactly divisible by 3, then the least value ofxis (a) 0 (b) 1 (c) 2 (d) 3 Solution: (b) 1 If a number is exactly divisible by 3, the sum of the digits must also be divisible by 3. $5+x+6=11+x$ must be divisible by 3 . The smallest value of $x$ is 1 . $x=1$ $\Rightarrow x+11=12$ is divisible by $3 .$...
Read More →Complete the magic square:
Question: Complete the magic square: Solution: The magic square is completed assuming that the sum of the row, columns and diagonals is 30. This is because the sum of all the number of the last column is 30....
Read More →Find the area of the shaded region given in figure
Question: Find the area of the shaded region given in figure Solution: Join JK, KL, LM and MJ, Their are four equally semi-circles and LMJK formed a square. Their are four equally semi-circles and $L M J K$ formed a square. $\therefore \quad F H=14-(3+3)=8 \mathrm{~cm}$ So the side of square should be $4 \mathrm{~cm}$ and radius of semi-circle of both ends are $2 \mathrm{~cm}$ each. $\therefore \quad$ Area of square $J K L M=(4)^{2}=16 \mathrm{~cm}^{2}$ Area of semi-circle $H J M=\frac{\pi r^{2}...
Read More →Fibonacci numbers Take 10 numbers as shown below:
Question: Fibonacci numbersTake 10 numbers as shown below: a,b, (a+b), (a+ 2b), (2a+ 3b), (3a+ 5b), (5a+ 8b), (8a+ 13b), (13a+ 21b), and (21a+ 34b). Sum of all these numbers = 11(5a+ 8b) = 11 7th number. Takinga= 8,b= 13; write 10 Fibonacci numbers and verify that sum of all these numbers = 11 7th number. Solution: Given: $a=8$ and $b=13$ The numbers in the Fibonnaci sequence are arranged in the following manner: $1 s t, 2 n d,(1 s t+2 n d),(2 n d+3 t h),(3 t h+4 t h),(4 t h+5 t h),(5 t h+6 t h)...
Read More →Fill in the numbers from 1 to 6 without repetition,
Question: Fill in the numbers from 1 to 6 without repetition, so that each side of the triangle adds up to 12. Solution: 6+2+4 = 124+3+5 = 126+1+5 = 12...
Read More →complete the magic square given below,
Question: complete the magic square given below, so that the sum of the numbers in each row or in each column or along each diagonal is 15. Solution: Taking the diagonal that starts with 6: $6+5+x=15 \Rightarrow x=4$ Now, taking the first row: $6+1+x=15 \Rightarrow x=8$ Taking the last column: $8+x+4=15 \Rightarrow x=3$ Taking the second column: $1+5+x=15 \Rightarrow x=9$ Taking the second row: $x+5+3=15 \Rightarrow x=7$ Taking the diagonal that begins with 8: $8+5+x=15 \Rightarrow x=2$...
Read More →The central angles of two sectors of circles
Question: The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120 and 40. Find the areas of the two sectors as well as the lengths of the corresponding arcs. What do you observe? Solution: Let the lengths of the corresponding arc be l1and l2 Given that, radius of sector $P O, O P=7 \mathrm{~cm}$ and radius of sector $\mathrm{AO}_{2} \mathrm{BA}=21 \mathrm{~cm}$ Central angle of the sector $P O, Q P=120^{\circ}$ and central angle of the sector $\mathrm{AO}_{2} \m...
Read More →Area of a sector of central angle 200°
Question: Area of a sector of central angle 200 of a circle is 770 cm2. Find the length of the corresponding arc of this sector. Solution: Let the radius of the sector AOBA be r. Given that, Central angle of sector $A O B A=\theta=200^{\circ}$ and area of the sector $A O B A=770 \mathrm{~cm}^{2}$ We know that, area of the sector $=\frac{\pi r^{2}}{360^{\circ}} \times \theta^{\circ}$ $\therefore \quad$ Area of the sector, $770=\frac{\pi r^{2}}{360^{\circ}} \times 200$ $\Rightarrow \quad \frac{77 ...
Read More →The length of the minute hand of a clock is 5 cm.
Question: The length of the minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6 : 05 am and 6 : 40 am Solution: We know that, in $60 \mathrm{~min}$, minute hand revolving $=360^{\circ}$ In 1 min, minute hand revolving $=\frac{360^{\circ}}{60^{\circ}}$ $\therefore \quad$ In $(6: 05$ am to $6: 40 \mathrm{am})=35 \mathrm{~min}$, minute hand revolving $=\frac{360^{\circ}}{60^{\circ}} \times 35=6 \times 35$ Given that, length of minute hand $(r)=5 \mathrm{~...
Read More →An archery target has three regions formed
Question: An archery target has three regions formed by three concentric circles as shown in figure. If the diameters of the concentric circles are in the ratio 1:2:3, then find the ratio of the areas of three regions. Solution: Let the diameters of concentric circles be k, 2k and 3k. $\therefore$ Radius of concentric circles are $\frac{k}{2}, k$ and $\frac{3 k}{2}$. $\therefore \quad$ Area of inner circle, $A_{1}=\pi\left(\frac{k}{2}\right)^{2}=\frac{k^{2} \pi}{4}$ $\therefore$ Area of middle r...
Read More →Find three whole numbers whose product and sum are equal.
Question: Find three whole numbers whose product and sum are equal. Solution: The three whole numbers are 1, 2 and 3. $1+2+3=6=1 \times 2 \times 3$...
Read More →All the vertices of a rhombus lie on a circle.
Question: All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is $1256 \mathrm{~cm}^{2}$, (use $\pi=3.14$ ) Solution: Let the radius of the circle be r. Given that. Area of the circle $=1256 \mathrm{~cm}^{2}$ $\pi r^{2}=1256$ $\Rightarrow$$r^{2}=\frac{1256}{\pi}=\frac{1256}{3.14}=400$ $\Rightarrow$ $r^{2}=(20)^{2}$ $\Rightarrow$ $r=20 \mathrm{~cm}$ $\therefore$ So, the radius of circle is $20 \mathrm{~cm}$. $\Rightarrow$ Diameter of circle $=2 \time...
Read More →Find two numbers whose product is a 1-digit
Question: Find two numbers whose product is a 1-digit number and the sum is a 2-digit number. Solution: 1 and 9 are two numbers, whose product is a single digit number. $\therefore 1 \times 9=9$ Sum of the numbers is a two digit number. $\therefore 1+9=10$...
Read More →Replace A, B, C by suitable numerals.
Question: Replace A, B, C by suitable numerals. Solution: $(A-4)=3 \Rightarrow A=7$ Also, $6 \times 6=36 \Rightarrow C=6$ $36-36=0 \Rightarrow B=6$ $\therefore A=7$ $B=C=6$...
Read More →Replace A, B, C by suitable numerals.
Question: Replace A, B, C by suitable numerals. Solution: $A \times B=B \Rightarrow A=1$ In the question: First digit = B+1 Thus, 1 will be carried from 1+B2and becomes (B+1) (B2-9) B. C = B2-1 Now, all B, B+1 and B2 -9 are one digit number. This condition is satisfied for B=3 or B=4. For B 3, B2-9 will be negative. For B3, B2-9 will become a two digit number. For B=3 , C = 32- 9 = 9-9 = 0 For B = 4, C = 42-9 = 16-9 = 7 Required answer: A=1, B=3, C = 0 or A=1, B=4, C = 7...
Read More →Replace A, B, C by suitable numerals.
Question: Replace A, B, C by suitable numerals. Solution: $(\mathrm{B} \times 3)=\mathrm{B}$ Then, B can either be 0 or 5 . If $\mathrm{B}$ is 5 , then 1 will be carried. Then, $\mathrm{A} \times 3+1=\mathrm{A}$ will not be possible for any number. $\therefore \mathrm{B}=0$ $\mathrm{A} \times 3=\mathrm{A}$ is possible for either 0 or 5 . If we take $\mathrm{A}=0$, then all number will become 0 . However, this is not possible. $\therefore \mathrm{A}=5$ Then, 1 will be carried. $\therefore \mathrm...
Read More →Replace A, B, C by suitable numerals.
Question: Replace A, B, C by suitable numerals. Solution: $5-A=9$ This implies that 1 is borrowed. We know: $15-6=9$ $\therefore A=6$ $B-5=8$ This implies that 1 is borrowed. $13-5=8$ But 1 has also been lent $\therefore B=4$ $C-2=2$ This implies that 1 has been lent. $\therefore C=5$ $\therefore A=6, B=4$ and $C=5$...
Read More →Replace A, B by suitable numerals.
Question: Replace A, B by suitable numerals. Solution: First look at the left column, which is: $6-A=3$ This implies that the maximum value of A can be 3. $A \leq 3$ ....(1) The next column has the following: $A-B=7$ To reconcile this with equation (1), borrowing is involved. We know: $12-5=7$ $\therefore A=2$ and $B=5$...
Read More →Replace A, B, by suitable numerals.
Question: Replace A, B, by suitable numerals. Solution: $A+A+A=A \quad$ (with 1 being carried over) This is satisfied if $A$ is equal to 5 . When $A=5$ : $A+A+A=15$ ( 1 is carried over) Or $B=1$ $\therefore A=5$ and $B=1$...
Read More →Replace A, B, C by suitable numerals.
Question: Replace A, B, C by suitable numerals. Solution: $A=7, A+6=7+6=13 \quad$ (1 is carried over) $(1+B+9)=17$, or $B=7 \quad$ ( 1 is carried over) $A=7, B=7$ and $C=4 \quad(1$ is carried over $)$ $\therefore A=7, B=7$ and $C=4$...
Read More →Replace A, B, C by suitable numerals.
Question: Replace A, B, C by suitable numerals. Solution: $A=6$ $\therefore \quad A+7=6+7=13$ 1 is carried over. $(1+5+8)=14$ 1 is carried over. $\therefore B=4$ and $C=1$ $\therefore A=6, B=4$ and $C=1$...
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