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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 66951. |
gIff()fog(seckftarik)X=than.9.Find the distance between the parallel line 3x-3y-4- 10x-6y9-0d0 and (2x-Dx-(8k-IDy-6 0 are |
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| 66952. |
)1/Find the circumference of the circles with the following radius: (Take π(a) 14 cm(b) 28 mm(c) 21 cm |
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Answer» 2*pie*ra)2×14×22/7=88cmb)28×2×22/7=176mmc)21×2×22/7=132cm |
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| 66953. |
● Complete the following table to draw the graph of 2x -6y = 3(r, y) |
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| 66954. |
The equation of the altitude of the AABC whosevertices are A(-4, 2); B(6, 5) and C(1, -4)can be:(A) 10x +3y +20 (B) 5x +9y +20 |
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| 66955. |
Solve the equation** - 10x + 26x2 - 10x + 1 = 0 |
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Answer» (x²-6x+1)(x²-4x²+1) is write the answer |
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| 66956. |
the following table shows the cumulative frequency distribution of height (in cm) of 100Height (in cm)Number of studentsAbove 120100Above 125Above 130Above 135Above 140Construct a frequency distribution table for the above data, |
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Answer» hii can you be my friend U can see that Number of students are in a Q.Therefore that is the value of CF for finding value of F u want to subtract. |
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| 66957. |
xfx and y are in direct proportion in the following table find y |
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Answer» x/y= x1/y1= x2/y2x/y= x1/y1y/x= y1/x1y/x*x1= y1 |
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| 66958. |
MPLE1 Show that -2,y 3 satisfy the linear equation 3x -4y+60UTION Substituting r? tu |
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Answer» If point x = 2, y = 3 satisfy equation 3x - 4y + 6 = 0 Then, for given value of x and y equation will be equal to zero Put value of x = 2, y = 3 in eqWe get, 3*2 - 4*3 + 6= 12 - 12= 0 Therefore given points satisfy linear equation |
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| 66959. |
(I) Complete the following table to draw the graph of 2x-6y 3.0(r, y) |
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| 66960. |
1) TULey20. Find the greatest number that divides 59 and 54leaving remainders 3 and 5, respectively.(a) 3(b) 7(c) 8(d) 5 |
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Answer» B) 7 is the right answer Seven is the right option. B)seven is the right answer 7 is right answer for your question |
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| 66961. |
2. TU JUULU3. What should be added to 1,80,78,975 to get 5,55,55,555? |
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Answer» Subtract 18078975 from 55555555 55,555,555−18,078,975=37476580so, 37476580 is added to 18078975 to get 55555555 |
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| 66962. |
TU vaiue 0TIf the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the valueof p. |
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| 66963. |
nensions of a cuboid are in the ratio of 1:2:3 and its total surface area is 88 m2. Findof tudimensiohe dimensionsper m. |
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Answer» ok I have got the answers |
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| 66964. |
B D^{2}=B C^{2}+C D^{2}=(12)^{2}+5^{2}=169 |
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Answer» BD*BD=BC*BC+CD*CD =12*12+5*5=144+25 =169=13*13 |
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| 66965. |
Find the sum by suitable rearrangement:238 + 695 + 162154 + 197 + 46+ 203 |
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Answer» The ans is 238+162+(695). =400+695. =1095 |
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| 66966. |
Find the sum by suitable rearrangement11. – 142+311+(-58)12.-142+425+75 |
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Answer» 11. -142+311+(-58) =311-142-58 =311-(142+58) =311-200 =11112. -142+425+75 =425+75-142 =500-142 =358 |
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| 66967. |
\sqrt{12^{3 m-2}}=12^{2}, \pi \times m= |
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| 66968. |
\frac { x ^ { 2 } + 12 x - 20 } { 3 x - 5 } = \frac { x ^ { 2 } + 8 x + 12 } { 2 x + 3 } |
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| 66969. |
ILLUSTRATION 1. A man buys a toy for 25 and sells it for30. Find his gainper cent.SOLUTION |
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Answer» According to the question. C.P = Rs. 25 S.P = Rs. 30 Profit = S.P - C.P = 30 - 25 = Rs. 5 P% =5/25×100=20% |
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| 66970. |
A man buys an article for Rs. 27.50 andsells it for Rs. 28.60. Then his gain per-cent is |
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| 66971. |
All questions are compulsory (edtiDivide rupee 351 into two parts such that onemay be other as 2:7.m among his children in th |
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| 66972. |
5x +3y = 2xy10x +9y = 5xy |
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| 66973. |
7. A man buys an article for Rs.27.50 and sells it for Rs.28.60. Find his gain percent?28 60 mm |
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Answer» plz I want answer Given : cp=27.50RsSp=28.60 Rsgain=Sp-cp=27.50-28.60=1.1 GAIN% = GAIN/ CP*100 = 1.1/27.50*100 = 4% |
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| 66974. |
Solve for x and y 5÷x+6y=13 3÷x +4y=7 |
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Answer» 5/x+6y= 13---(1)3/x+4y= 7---(2)multiple by 2 in equation 1 3 in eq 210/x+12y= 269/x+12y= 21by subtracting 10/x-9/x= 26-2110-9/x= 55x= 1x= 1/5putting the value of X in eq 15/x+6y= 135/1/5+6y= 1325+6y=136y= 13-25=-12y= -12/6= -2 5/x+6y=13; 3/x+4y=7; 1/x=k; 5k+6y=13; 3k+4y=7; 3(5k+6y=13); 15k+18y=39; 5(3k+4y=7); 15k+20y=35/ 2y=4; y=2; 5k+6(2)=13; 5k=13-12=1; k=1/5=x |
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| 66975. |
\operatorname { sin } 78 ^ { \circ } \operatorname { cos } 18 ^ { \circ } - \operatorname { cos } 78 ^ { \circ } \operatorname { sin } 18 ^ { \circ } |
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| 66976. |
.Factorise:Write the following cubes in expanded form: |
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| 66977. |
Factorise 6xy - 4y + 6 â 9y, |
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| 66978. |
30. Fill in the blanks to make the given expression a perfect square :(i) 160.2 +9b2 +(iii) 4a2 + 20a0+(ii) 25 Îą2 + 16 b2-...(iv) 9a -24ab +2 |
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| 66979. |
TU3. Express each of the following decimals in the form where p, q areintegers and q40(1) 0.2 (2014) (1) 0,53 (2010)(iii) 2.9312010, '151(iv) 18.48 12014) (v) 0.235 (2010) (vi) 0.0032 |
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| 66980. |
Complete the following table to draw the graph of 2x-6y=3 |
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| 66981. |
2. Add the following using suitable rearrangement.0 3.2.1os84 |
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Answer» -3/4+7/8+1/4=-6/8+7/8+2/8=3/8 |
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| 66982. |
A man buys toffees at 10 for ₹3 and sells them at 8 for ₹3,find his gain percent |
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| 66983. |
A man buys toffees at 10 for ₹3 and sells them at 8 for ₹ 3, find his gain percent. |
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Answer» CP of 10 toffees =Rs. 3CP of 1 toffee =10/3=Rs. 3.33SP of 8 toffees =Rs. 3SP of 1 toffee =8/3 =Rs. 2.66Since SP <CP, therefore LossLoss = CP-SP=3.33-2.66=Rs. 0.67Loss % =Loss/CP×100=0.67/3.33×100=20.12% |
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| 66984. |
hs.A man buys toffees at10 for3 and sells them at 8 for3, find his gain per cent. L |
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| 66985. |
14.1(6) 12x, 36(iv) 2x, 3x2, 4(vi) 16x3-4x2, 32x() 2y,22xy () 14 pq, 28p'q(v) 6 abc, 24ab?, 12 a'b(vi) 10 pq, 20qr, 30rp2 122. |
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Answer» thanks |
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| 66986. |
torise:(i) 12x +15(i) 16a? - 24ab(i) 24 x 8 - 36x’y(i) 9x - 6x2 +12x(i) 14 x3 +21x*y - 28x(i) x(x + 3) +5(x +3)1.) ha 2) |
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| 66987. |
My a03x—9y—-2=0 |
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| 66988. |
- w3k 351 Y:M/X | ) [| | |
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Answer» 10*10=100*110=1100015*15=225*500=112500 |
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| 66989. |
13. What should be added to -351 to obtain-53 |
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Answer» Let the number to be added be x.-351+x=-53 x=351-53=298 So 298 should be added to -351 to get -53. |
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| 66990. |
Q40 78% of 450 is:(A) 351(C) 296(B) 312(D) 303 |
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| 66991. |
of the rectangle(ii) Find the ratio of the perimeter to the area of the giventriangle.5235110210th Standard Mathematics |
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| 66992. |
3y(y+3)+6y(9y+3) |
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Answer» 3y(y+3)+6y(9y+3)=3y²+9y+54y²+18y=57y²+27y=y(57y+27) 3y(y+3)+6y(9y+3=3y^2+9y+54y^2+18y=57y^2+27y=y(57y+27) {^=power} 3y(y+3)+6y(9y+3)=3y^2+9y+54y^2+18y=57y^2+27y=3y(19y+9) y(57y+27) is the correct answer of the given question 3y ( 19 y + 9) is the answer 3y(y+3)+6y(9y+3)3y^2+9y+54y^2+18yCOMBINING LIKE TERMS,3y^2+54y^2+9y+18y57y^2+27yy(57y+27) solution:3y^2+9y+54y^2+18y =(3y^2+54y^2)+(9y+18y)=57y^2+27y=y(57y+27) |
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| 66993. |
6.Write the following cubes in expanded form:0 (2r+1) |
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Answer» thanks a lot this helped me!! |
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| 66994. |
following cubes in expandex + 1 ) ^3(ii) (2a-3b)te the following using suitablether+1)3 |
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| 66995. |
Subtract 24ab- 10b- 18a from 30ab + 12b+ 14a. |
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| 66996. |
7asquare + 14a |
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Answer» Factorization 7a*a + 14a = 7a(a+2) If you find this answer helpful then like it. |
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| 66997. |
.9. A fauit Vendos buys oranges at fam0 and anl equal number a 1O fo 20.He mixes them and sells at 21.60 oF persdozen. find his gain e |
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Answer» NO OF ORANGES HE BOUGHT =6+10=16TOTAL COST=10+20=RS.30C.P OF EACH ORANGES IS 30/16NOW HE SELLS IT AT RS.21.6 FOR 12S.P OF ORANGES=21.6/12SINCE 30/16 IS GRATER THAT 21.6/12HE HAD A LOSSTHE LOSS =30/16-21.6/12=RS 3.6 thanks |
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| 66998. |
Find the zeroes of the fothe zeroes and the coe(i) x^2 -2x -8 |
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Answer» x² -2x -8x² +2x -4x -8x(x+2) -4(x+2)(x-4)(x+2)x = 4, -2 |
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| 66999. |
A man buys 10 eatables for? 3 and sells 8 of them for3: find his gain per cent. |
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| 67000. |
13. Subtract 24ab - 10b - 18a from 30ab + 12b + 14a. |
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Answer» 6ab+22b+32a is the right answer 6ab+22b+32a is the correct answer of the given question as the answer is as follow6ab+22b+32a 6ab+22b+32a is correct answer. |
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